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ELEMENTS OF 

6 

PLANE AND SOLID 

GEOMETRY. 

TOGETHER WITH THE 

ELEMENTS OF PLANE AND SPHEEICAL 

TRIGONOMETRY, 

AND AN ARTICLE ON INVERSE TRIGONOMETRICAL 

FUNCTIONS. 



GERARDUS BEEKMAN DOCHARTT, LL.D., 

PEOFE5SOS OF MATHEMATICS EM THE NEW YOBK FBEE ACADEMY, AND AUTHOR 
A "PRACTICAL AND COMMESCIAL ARITHMETIC," AND THE 

"institutes of algebra." 



SIXTH EDITION, VERY GREATLY IMPROVED AND ENLARGED. 



** 



NEW YORK: 

HARPER & BROTHERS, PUBLISHERS, 

FRANKLIN SQUARE, 

1862. 



jUt.J^Sfct^ 



4 



y 



^ 



Entered, according to Act of Congress, in the year one thousand 
eight hundred and sixty - two, by 

Harper & Brothers, 

in the Clerk's Office of the District Court of the Southern District 
of New York. 



I t 1 2 £ 



2 ■- \ 



o 

X 



TO 



HORACE WEBSTER, LL.D., 

PEESIDEXT OF THE FACULTY OF THE NEW YOKE FEEE ACADEMY, 

AS A TRIBUTE OF RESPECT FOR THE FAITHFUL PERFORM- 
ANCE OF THE DUTIES BELONGING TO HIS OFFICE, 



A TOKEN OF SINCERE REGARD FOR HIS DEVOTION TO THE 
CAUSE OF SOUND EDUCATION, AND 

A MEMORIAL OF PERSONAL FRIENDSHIP CEMENTED BY A 
DAILY INTERCOURSE OF MANY YEARS, 

®l)ig ttolmne 

IS MOST SINCERELY DEDICATED BY 

THE AUTHOR. 



PREFACE. 



In this volume I have endeavored to furnish the 
student with every aid which can be afforded con- 
sistently with securing the end which mathematical 
studies are designed to attain. I have not attempted 
to lessen the amount of thought and attention which 
are necessary for him to put forth, but only to prevent 
him from wasting that thought and attention. 

The most perfect demonstration of a mathematical 
theorem is that which, while rigorously exact, is ex- 
pressed in the fewest words. The " Elements of Eu- 
clid" are acknowledged to be absolutely perfect in this 
respect. Taking these for a basis and model, I have 
brought the essential features of Geometry, Trigonom- 
etry, and Mensuration within the compass of a single 
volume, thus enabling the student to remember all 
the propositions, and the dependence of each upon 
those which precede it. 

In the preparation of this volume I have made free 
use of whatever I have found in the works of French 
and English mathematicians which was adapted to 
my purpose. In the chapters on Trigonometry and 
Mensuration the student will find much that will, by 
exercising his analytical powers, fit him for the suc- 
cessful study of the higher branches of mathematics. 



VI PREFACE. 



I have prepared this treatise with a special view to 
meet the want of students in the New York Free 
Academy, who are early trained to appreciate the re- 
fined and subtle method of demonstration which is 
occasionally used in this volume. I trust, however, 
that this work will be found useful in other institu- 
tions in which mathematical studies are allowed to 
have their proper place. Gr. B. D. 

New York Free Academy, September 1, 1862. 



CONTENTS. 



PAGE 

Book I. Definitions 1 

Book II. Of the Circle 34 

Book III. OfKatios and Proportions 49 

Book IV 56 

Problems 69 

Book V. Geometry of Planes 91 

Book VI. Solid Angles 103 

Book VII. Solid Geometry 109 

Book VIII. Spherical Geometry ,. 123 

Elements of Plane Trigonometry 133-166 

Inverse Trigonometrical Functions ^ 16Q 

Solution of Right-angled Plane Triangles by means of Loga- 
rithms 169 

Solution of Oblique-angled Plane Triangles 171 

Mensuration 174 

Mensuration of Surfaces 174 

Mensuration of Solids 184 

Surfaces of Solids 184 

Volume of Solids 189 

Spherical Trigonometry 197-211 

Right-angled Spherical Triangles 212 

Solution of Right-angled Spherical Triangles 217 

Solution of Oblique-angled Spherical Triangles 219 



GEOMETRY. 



BOOK I. 

DEFINITIONS. 

Geometry is a science which treats of the properties and 
relations of quantities or magnitudes having extension. 

1. A Point is that which has position, but no magni- 
tude nor dimensions ; neither length, breadth, nor thick- ■ 
ness. 

2. A Line is length without breadth or thick- 

ness. 

3. A Surface or Superficies is an extension 
or a figure of two dimensions, length and 
breadth ; but without thickness. 

4. A Body or Solid is a figure of three di- 
mensions, namely, length, breadth, and depth, or 
thickness. 



5. Lines are either Eight or Curved, or Mixed of these 
two. 

6. A Eight Line or Straight Line lies all 

in the same direction, between its extremi- — 

ties, and is the shortest distance between two 

points. From one point to another only one straight line 

can be drawn. 

"When a Line is mentioned simply, it means a Eight Line. 

7. A Curve continually changes its direction 
between its extreme points. 

8. Lines are either Parallel, Oblique, Perpendicular, or 
Tangential. 

9. Parallel Lines are always at the same 2 ' 
perpendicular distance ; and they never meet, 
though ever so far produced. 

10. Oblique Lines change their distance, and 
would meet if produced on the side of the least 
distance. 

A 







\, 




GEOMETRY. 



11. One line is Perpendicular to another 
when it inclines not more on the one side than 
the other, or when the angles on both sides of 
it are equal. 

12. A Line or Circle is Tangential, or is a 
Tangent to a Circle or other curve, when it 
touches it without cutting, although both are 
produced. 

13. An Angle is the inclination or opening 
of two lines having different directions and 
meeting in a point. 

14. Angles are Right or Oblique, Acute or Obtuse. 

15. A Right Angle is that which is made 
by one line perpendicular to another. Or, 
when the angles on each side are equal to 
one another, they are right angles. 





16. An Oblique Angle is that which is made by two 
oblique lines, and is either less or greater than a right 
angle. 

17. An Acute Angle is less than a right angle. 

18. An Obtuse Angle is greater than a right 
angle. 

19. Superficies are either Plane or Curved. 

20. A Plane Superficies, or a Plane, is that with which 
a right line may every way coincide. Or, if the line touch 
the plane in two points, it will touch it in every point. But 
if not, it is curved. 

21. Plane figures are bounded either by right lines or 
curves. 

22. Plane figures that are bounded by right lines have 
names according to the number of their sides or of their 
angles ; for they have as many sides as angles ; the legist 
number being three. 

23. A figure of three sides and angles is called a Trian- 
gle ; and it receives particular denominations from the re- 
lations of its sides and angles. 



24. An Equilateral Triangle is that whose 
three sides are all equal. 





DEFINITIONS. 

25. An Isosceles Triangle is that which has 
two sides equal. 

26. A Scalene Triangle is that whose three sidelTare all 
unequal. 

27. A Eight -angled Triangle is that which has one right 
angle. 

28. Other Triangles are Oblique-angled, and are either 
obtuse or acute. 

29. An Obtuse-angled Triangle has one obtuse angle. 

30. An Acute-angled Triangle has all its three angles 
acute. 

31. A figure of Four sides and angles is called a Quad- 
rangle, or a Quadrilateral. 

32. A Parallelogram is a quadrilateral which has both 
its pairs of opposite sides parallel. And it takes the follow- 
ing particular names, viz., Rectangle, Square, Rhombus- 
Rhomboid. 

33. A Rectangle is a parallelogram having a 
right angle. 

34. A Square is an equilateral rectangle, hav- 
ing its length and breadth equal, or all its sides 
equal, and all its angles equal. 

35. A Rhomboid is an oblique-angled 
parallelogram. 

36. A Rhombus is an equilateral rhom 
boid, having all its sides equal, but its an 
gles oblique. 

37. A Trapezium is a quadrilateral which has not its op- 
posite sides parallel. 

38. A Trapezoid has only one pair of opposite sides par- 
allel. 

39. A Diagonal is a line joining any two opposite angles 
of a quadrilateral. 

40. Plane figures that have more than four sides are, in 
general, called Polygons ; and they receive other particular 
names, according to the number of their sides or angles. 
Thus, 

41. A Pentagon is a polygon of five sides; a Hexagon, 
of six sides ; a Heptagon, seven ; an Octagon^ eight ; a Non= 




4 GEOMETRY. 

agon, nine ; a Decagon, ten ; an Undecagon, eleven ; and a 
Dodecagon, twelve sides. 

42. A Regular Polygon has all its sides and all its angles 
equal. If they are not both equal, the polygon is Irregular. 

43. An Equilateral Triangle is also a Regular Figure of 
three sides, and the Square is one of four : the former being 
also called a Trigon, and the latter a Tetragon. 

44. Any figure is equilateral when all its sides are equal : 
and it is equiangular when all its angles are equal. When 
both these are equal, it is a regular figure. 

45. A Circle is a plane figure bounded by a 
curve line, called the Circumference, which is every 
where equidistant from a certain point within, call- 
ed its Centre. 

The circumference itself is often called a circle, and also 
the Periphery. 

46. The Radius of a circle is a line drawn 
from the centre to the circumference. 



47. An Arc of a circle is any part of the cir- 
cumference. 



iO 



48. A Chord is a right line joining the extrem- 
ities of an arc. 



s 



49. The Diameter of a circle is the chord which 
passes through the centre. 

50. A Segment is any part of a circle bounded r" "\ 
by an arc and its chord. \ i 

51. A Semicircle is half the circle, or a seg- 
ment cut off by a diameter. 

The half circumference is sometimes called the 
Semicircle. 

52. A Sector is any part of a circle which is 
bounded by an arc and two radii drawn to its ex- 
tremities. 





DEFINITIONS. 

53. A Quadrant, or Quarter of a circle, is a 
sector having a quarter of the circumference for 
its arc, and its two radii are perpendicular to each 
other. A quarter of the circumference is some- 
times called a Quadrant. 

54. The Height or Altitude of a figure is a 
perpendicular let fall from an angle, or its ver- 
tex, to the opposite side, called the base. 

55. In a Eight-angled Triangle, the side opposite the right 
angle is called the Hypothenuse ; and the other two sides are 
called the Legs, and sometimes the Base and Perpendicular. 

56. When an Angle is denoted by three letters, of which 
one stands at the angular point, and the other two on the 
two sides, that which stands at the angular point is read in 
the middle. 

57. The Circumference of every circle is supposed to be 
divided into 360 equal parts, called Degrees ; and each de- 
gree into 60 Minutes, each minute into 60 Seconds, and so 
on. Hence a semicircle contains 180 degrees, and a quad- 
rant 90 degrees. 

58. The Measure of an angle is an arc of any circle con- 
tained between the two lines which form that angle, the an- 
gular point being the centre ; and it is estimated by the 
number of degrees contained in that arc. 

59. Lines or Chords are said to be equidistant from the 
centre of a circle when j>erpendiculars drawn to them from 
the centre are equal. 

60. And the right line on which the Greater Perpendic- 
ular falls is said to be farther from the centre. 

61. An Angle in a Segment is that which is 
contained by two lines drawn from any point in 
the arc of the Segment to the two extremities of 
that arc. 

62. An Angle on a segment or an arc is that which is 
contained by two lines drawn from any point in the oppo- 
site or supplementary part of the circumference to the ex- 
tremities of the arc, and containing the arc between them. 

63. An Angle at the circumference is that 
whose angular point or summit is any where in 
the circumference. And an angle at the centre 
is that whose angular point is at the centre. 





GEOMETRY. 




64. A Right-lined figure is Inscribed in a cir- 
cle, or the circle Circumscribes it, when all the 
angular points of the figure are in the circumfer- 
ence of the circle. 

65. A Right-lined figure Circumscribes a cir- 
cle, or the circle is Inscribed in it, when all the 
sides of the figure touch the circumference of the 
circle. 

66. One Right-lined figure is Inscribed in an- 
other, or the latter Circumscribes the former, 
when all the angular points of the former are 
placed in the sides of the latter. 

67. A Secant is a line that cuts a circle, 
lying partly within, and partly without it. 



68. Two Triangles, or other right-lined figures, are said 
to be mutually equilateral when all the sides of the one are 
equal to the corresponding sides of the other, each to each ; 
and they are said to be mutually equiangular when the 
angles of the one are respectively equal to those of the 
other. 

69. Identical figures are such as are both mutually equi- 
lateral and equiangular, or that have all the sides and all 
the angles of the one respectively equal to all the sides and 
all the angles of the other, each to each ; so that if the one 
figure were applied to, or laid upon the other, all the sides 
of the one would exactly fall upon and cover all the sides of 
the other, the two becoming, as it were, but one and the 
same figure. 

70. Similar figures are those that have all the angles of 
the one equal to all the angles of the other, each to each, 
and the sides about the equal angles proportional. 

71. The Perimeter of a figure is the sum of all its sides 
taken together. 

72. A Proposition is something which is either proposed 
to be done or to be demonstrated, and is either a problem or 
a theorem. 

73. A Problem is something proposed to be done. 

74. A Theorem is something proposed which requires a 
demonstration. 



AXIOMS. 7 

75. A Lemma is something which is premised or demon- 
strated, in order to render what follows more easy. 

76. A Corollary is a consequent truth, gained immedi- 
ately from some preceding truth or demonstration. 

77. A Scholium is a remark or observation made upon 
something going before it. 



AXIOMS. 

1. Things which are equal to the same thing are equal to 
each other. 

2. "When equals are added to equals, the wholes are equal. 

3. When equals are taken from equals, the remainders 
are equal. 

4. When equals are added to unequals, the wholes are 
unequal. 

5. When equals are taken from unequals, the remainders 
are unequal. 

6. Things which are double of the same thing, or equal 
things, are equal to each -other. 

7. Things which are halves of the same thing are equal. 

8. Every whole is equal to all its parts taken together, or 
it is greater than any of its parts. 

9. Things which coincide, or fill the same space, are iden- 
tical, or mutually equal in all their parts. 

10. All right angles are equal to one another. 

11. Angles that have equal measures, or arcs, are equal. 



THEOREM I. 

If two triangles have two sides and the included angle in the 
one equal to two sides and the included angle in the other, the 
triangles will be identical, or equal in all respects. 

In the two triangles ABC, 
DEF, if the side AC be equal 
to the side DF, and the side 
BC equal to the side EF, and 
the angle C equal to the an- 
gle F, then will the two tri- 
angles be identical, or equal 
in all respects. 




8 GEOMETRY. 

For, conceive the triangle ABC to be applied to, or placed 
on, the triangle DEF, in such a manner that the point C 
may coincide with the point F, and the side AC with the 
side DF, which is equal to it. 

Then, since the angle F is equal to the angle C (by hyp.), 
the side BC will fall on the side EF. Also, because AC is 
equal to DF, and BC equal to EF (by hyp.), the point A 
will coincide with the point D, and the point B with the 
point E ; consequently the side AB will coincide with the 
side DE. Therefore the two triangles are identical, and 
have all their other corresponding parts equal (ax. 9), name- 
ly, the side AB equal to the side DE, the angle A to the 
angle D, and the angle B to the angle E. Q. E. D. 

THEOREM II. 

When two triangles have two angles and the included side in 
the one equal to two angles and the included side in the other, 
the triangles are identical, or have their other sides and angles 
equal. 

Let the two triangles ABC, 
DEF, have the angle A equal 
to the angle D, the angle B 
equal to the angle E, and the 
side AB equal to the side DE ; 
then these two triangles will 
be identical. 

For, conceive the triangle ABC to be placed on the tri- 
angle DEF in such manner that the side AB may fall ex- 
actly on the equal side DE. Then, since the angle A is 
equal to the angle D (by hyp.), the side AC must fall on the 
side DF ; and, in like manner, because the angle B is equal 
to the angle E, the side BC must fall on the side EF. Thus 
the three sides of the triangle ABC will be exactly placed 
on the three sides of the triangle DEF ; consequently the 
two triangles are identical (ax. 9), having the other two 
sides AC, BC, equal to the two DF, EF, and the remaining 
angle C equal to the remaining angle F. Q. E. D. 




THEOREMS. 




THEOREM m. 

In an isosceles triangle, the angles at the base are equal. Or, 
if a triangle have two sides equal, their opposite angles will also 
be equal. 

If the triangle ABC have the side AC 
equal to the side AB, then will the angle B 
be equal to the angle C 

For, conceive the angle A to be bisected, 
or divided into two equal parts, by the line 
AD, making the angle BAD equal to the an- 
gle CAD. B 

Then the two triangles ACD, ABD, have two sides and 
the contained angle of the one, equal to two sides and the 
contained angle of the other, viz., the side AC equal to AB, 
the angle BAD equal to CAD, and the side AD common ; 
therefore these two triangles are identical, or equal in all re- 
spects (th. 1) ; and, consequently, the angle C equal to the 
angle B. Q. E. D. 

Corol. 1. Hence the line which bisects the vertical angle 
of an isosceles triangle bisects the base, and is also perpen- 
dicular to it. 

Corol. 2. Hence, too, it appears that every equilateral tri- 
angle is also equiangular, or has all its angles equal. 



THEOREM IV. 

When a triangle has two of its angles equal, the sides oppo- 
site to them are also equal. 

Let ABC be a triangle, having the angle 
B equal to the angle C ; then will the side 
AC be equal to the side AB. 

For, if they are not equal, suppose AB> 
AC, and take BD=AC, and join DC 

Then, in the two triangles BCD, ABC, we 
have the side BD=:AC, by supposition; the 
side BC common, and the included angle B c 

CBD of the one triangle equal to the included angle BCA 
of the other triangle. Therefore the triangle BDC is equal 

A2 




10 GEOMETRY. 

to the triangle ABC (th. 1). That is, a part is equal to 
the whole, which is impossible (ax. 8). 

Hence there is no inequality of the sides AB and AC ; 
that is, they are equal, and the triangle is isosceles. Q. E. D. 

Corollary. Hence every equiangular triangle is also equi- 
lateral. 

THEOREM V. 

When two triangles have all the three sides in the one equal 
to all the three sides in the other, the triangles are identical, or 
have also their three angles equal, each to each. 

Let the two triangles ABC, ABD, c 

have their three sides respectively equal, ^-"""^V 

viz., the side AB equal to AB, AC to ^^^ \ ^ -r 
AD, and BC to BD ; then shall the two A \^^ | / 

triangles be identical, or have their an- ^^^J/ 

gles equal, viz., those angles that are op- 
posite to the equal sides ; that is, the angle BAC to the an- 
gle BAD, the angle ABC to the angle ABD, and the angle 
C to the angle D. 

For, conceive the two triangles to be joined together by 
their longest equal sides, and draw the line CD. 

Then, in the triangle ACD, because the side AC is equal 
to AD (by hyp.), the angle ACD is equal to the angle ADC 
(th. 3). In like manner, in the triangle BCD, the angle 
BCD is equal to the angle BDC, because the side BC is equal 
to BD. Hence, then, the angle ACD being equal to the an- 
gle ADC, and the angle BCD to the angle BDC, by equal 
additions the sum of the two angles ACD, BCD, is equal 
to the sum of the two ADC, BDC (ax. 2) ; that is, the whole 
angle ACB equal to the whole angle ADB. 

Since, then, the two sides AC, CB, are equal to the two 
sides AD, DB, each to each (by hyp.), and their contained 
angles ACB, ADB, also equal, the two triangles ABC, 
ABD, are identical (th. 1), and have the other angles equal, 
viz., the angle BAC to the angle BAD, and the angle ABC 
to the angle ABD. Q. E. J). 



THEOREMS. 11 



THEOREM VI. 

When one line meets another, the angles which it makes on 
the same side of the other are together equal to two right angles. 

Let the line AB meet the line CD ; 
then will the two angles ABC, ABD, 
taken together, be equal to two right an- 



For, first, when the two angles ABC, 
ABD, are equal to each other, they are c B t> 

both of them right angles (def. 15). 

But when the angles are unequal, suppose BE drawn per- 
pendicular to CD. Then, since the two angles EBC, EBD, 
are right angles (def. 15), and the angle EBD is equal to 
the two angles EBA, ABD, together (ax. 8), the three an- 
gles EBC, EBA, and ABD are equal to two right angles. 

But the two angles EBC, EBA, are together equal to the 
angle ABC (ax. 8). Consequently, the two angles ABC, 
ABD, are also equal to two right angles. Q. E. D. 

Corol. 1. Hence also, conversely, if the two angles ABC, 
ABD, on both sides of the line AB, make up together two 
right angles, then CB and BD form one continued right 
line CD. 

Corol. 2. Hence all the angles which can be made, at any 
point B, by any number of lines, on the same side of the 
right line CD, are, when taken all together, equal to two 
right angles. 

Corol. 3. And as all the angles that can be made on the 
other side of the line CD are also equal to two right angles, 
therefore all the angles that can be made quite round a 
point B, by any number of lines, are equal to four right 
angles. 

Corol. 4. Hence, also, the whole circumference 
of a circle, being the sum of the measures of all 
the angbs that can be made about the centre F 
(def. 57), is the measure of four right angles. 
Consequently, a semicircle, or 180 degrees, is the measure 
of two right angles ; and a quadrant, or 90 degrees, the 
measure of one right angle. 




12 GEOMETRY. 



THEOREM VH. 

When two lines intersect each other, the opposite angles are 
equal. 

Let the two lines AB, CD, intersect 
in the point E ; then will the angle 
AEC be equal to the angle BED, and 



the angle AED be equal to the angle A E\ 33 

CEB. \ 

For, since the line CE meets the line ^ 

AB, the two angles AEC, BEC, taken together, are equal 
to two right angles (th. 6). 

In like manner, the line BE, meeting the line CD, makes 
the two angles BEC, BED, equal to two right angles. 

Therefore, the sum of the two angles AEC, BEC, is equal 
to the sum of the two BEC, BED (ax. 1). 

And if the angle BEC, which is common, be taken away 
from both these, the remaining angle AEC will be equal to 
the remaining angle BED (ax. 3). 

And, in like manner, it may be shown that the angle AED 
is equal to the opposite angle BEC. 

THEOREM Vm. 

When one side of a triangle is produced, the exterior angle 
is greater than either of the two interior opposite angles. 

Let ABC be a triangle, having the side 
AB produced to D ; then will the out- 
ward angle CBD be greater than either 
of the inward opposite angles A or C 

For, conceive the side BC to be bisect- 
ed in the point E, and draw the line AE, 
producing it till EF be equal to AE ; 
and join BF. 

Then, since the two triangles AEC, 
BEF, have the side AE=the side EF, and the side CE=the 
side BE (by suppos.), and the included or opposite angles at 
E also equal (th. 7), therefore those two triangles are equal in 
all respects (th. 1), and have the angle C = the correspond- 
ing angle EBF. But the angle CBD is greater than the 





THEOREMS. 13 

angle EBF ; consequently, the said outward angle CBD is 
also greater than the angle C. 

In like manner, if CB be produced to Gr, and AB be bi- 
sected, it may be shown that the outward angle ABG, or 
its equal CBD, is greater than the other angle A. 

THEOREM IX. 

The greater side of every triangle is opposite to the greater 
angle, and the greater angle opposite to the greater side. 

Let ABC be a triangle, having the side AB A^ 
greater than the side BC ; then will the an- 
gle ACB, opposite the greater side AB, be 
greater than the angle A, opposite the less 
side CB. 

For, on the greater side AB, take the part 
BD equal to the less side BC, and join CD. c B 

Then, since ACD is a triangle, the exterior angle BDC is 
greater than the interior opposite angle A (th. 8). But the 
angle BCD is equal to the §aid exterior angle BDC, be- 
cause BD is equal to BC (th. 3). Consequently, the angle 
BCD also is greater than the angle A. And since the an- 
gle BCD is only a part of ACB, much more must the whole 
angle ACB be greater than the angle A. Q. E. D. 

Again, conversely, if the angle C be greater than the an- 
gle A, then will the side AB, opposite the former, be greater 
than the side BC, opposite the latter. 

For, if AB be not greater than BC, it must be either 
equal to it, or less than it. But it can not be equal, for 
then the angle C would be equal to the angle A (th. 3), 
which it is not by the supposition. Neither can it be less, 
for then the angle C would be less than the angle A, by the 
former part of this, which is also contrary to the supposi- 
tion. The side AB, then, being neither equal to AC nor 
less than it, must necessarily be greater. Q. E. D. 

THEOREM X. 

The sum of any two sides of a triangle is greater than the 
third side. * 



Let ABC be a triangle : then will the sum of two of its 



14 GEOMETRY. 

sides, as AB+BC, be greater than the third 
side AC. 

For the straight line AC is the shortest -^ 
distance between the points A and C (def. 6); hence AB+ 
BC>AC. 

Corol. If from both members of the inequality we subtract 
the side BC, we shall have 

AB>AC-BC: 
that is, the difference of any two sides of a triangle is less than 
the third side. 

THEOREM XI. 

If from a point ivithin a triangle tivo lines be drawn to the 
extremities of either side, their sum will be less than the sum of 
the other two sides of the triangle. 

Let ABC be a triangle, and D any 
point within it. Draw CD, DB, to the 
extremities of the side BC ; then will 
BD+CD be less than AB-f AC. 

Produce BD to E ; then BA-f-AE> 
BE. To both members of this inequal- 
ity add EC; then will BA+AE+EC 
>BE+EC ; or, which is the same thing, BA-f AC>BE 
+EC. 

Again, in the triangle DEC, we have 
DE+EC>DC. 
To both members of this inequality add BD ; then DE-j-BD 
+EC>DC+BD, or 

BE+EC>BD-fDC. 
Hence, much more is 

BA+AC>BD+DC, or 
BD+DC<BA+AC 

THEOREM XH. 

If two triangles have two sides of the one equal to two sides 
of the other, but the included angles unequal, the third sides ivill 
be unequal, and the greater will be in that triangle which has 
the greater included angle. 

Let ABC, DEF be two triangles in which AB^DE, AC 





THEOREMS. 15 

= DF, BAC < EDF. A 
Then will EF>BC. 

For at the point D 
make the angle EDG 
equal to the angle 
BAC ; take DG equal 

to AC, and join GE. B C ^-^G 

Then will the triangle DEG equal the triangle ABC (th. 
1), and EGzzzBC. 

But EG<EI+IG, and DF<DI+IF (Prop. 10). 

By addition of these inequalities, EG-J-DF<EI+IF + 
DI+IG; or, EG+DF<EF+DG. 

Taking away the equals DF and DG from the members 
of the last inequality, there remains EG<EF. But EG= 
BC.\BC<EF. Q. E. D. 

If the point G fall within instead of without the triangle 
DEF, we should have DG+GE<DF+EF (th. 11) ; and, 
taking away the equals DG and DF, there remains EG< 
EF. If the point G fall on EF, the theorem is evident. 

The converse of this proposition is also true, viz., that if 
two sides of one triangle be equal to two sides of another, 
and the third sides unequal, the angle opposite the smaller 
third side will be less than the one opposite the larger. For 
if the angle were greater, by the above proposition the third 
side must be greater ; and if it were equal, it must (by th. 
1) be equal. But the third side is neither greater nor equal; 
therefore the angle opposite, being neither greater than nor 
equal to the angle of the other triangle, must be less. 

THEOREM XTTT. 

When a line intersects two parallel lines, it makes the alter- 
nate angles equal to each other. 

Let the line EF cut the two parallel lines 
AB, CD ; then will the angle AEF be 
equal to the alternate angle EFD. -q 

For if they are not equal, one of them 
must be greater than the other ; let it be EFD, for instance, 
which is the greater, if possible ; and conceive the line FB 
to be drawn, cutting off the part or angle EFB equal to the 
angle AEF, and meeting the line AB in the point B. 




16 GEOMETRY. 

, Then, since the exterior angle AEF, of the triangle BEF, 
is greater than the interior opposite angle EFB (th. 8) ; and 
since these two angles also are equal (by the constr.), it fol- 
lows that those angles are both equal and unequal at the 
same time, which is impossible. Therefore the angle EFD 
is not unequal to the alternate angle AEF, that is, they are 
equal to each other. Q. E. D. 

Corol. Right lines which are perpendicular to one of two 
parallel lines, are also perpendicular to the other. 

THEOREM XIV. 

When a line, cutting two other lines, makes the alternate an- 
gles equal to each other, those two lines are parallel. 

Let the line EF, cutting the two lines AB, a e/ B 
CD, make the alternate angles AEF, DFE, / 

equal to each other ; then will AB be par- c /^^~^_ ** 
allel to CD. G^ 

For, if they be not parallel, let some other line, as FG-, be 
parallel to AB. Then, because of these parallels, the angle 
AEF is equal to the alternate angle EFG (th. 13). But 
the angle AEF is equal to the angle EFD (by hyp.). There- 
fore the angle EFD is equal to the angle EFG (ax. 1) ; that 
is, a part is equal to the whole, which is impossible. There- 
fore no line but CD can be parallel to AB. Q. E. D. 

Gorol. Those lines which are perpendicular to the same 
line, are parallel to each other. 

THEOREM XV. 

When a line cuts two parallel lines, the exterior angle is equal 
to the interior opposite one on the same side ; and the two interi- 
or angles on the same side are together equal to two right angles. 

Let the line EF cut the two E 

parallel lines AB, CD ; then will / 

the exterior angle EGB be equal G-/ 

to the interior opposite angle -°- ~T 

GHD, on the same side of the -/ 

line EF; and the two interior q ^jL. D 

angles BGH, GHD, taken to- /H 

gether, will be equal to two right / 

angles. ^ 



THEOREMS. 17 

For, since the two lines AB, CD, are parallel, the angle 
AGH is equal to the alternate angle GHD (th. 13). But 
the angle AGH is equal to the opposite angle EGB (th. 7). 
Therefore the angle EGB is also equal to the angle GHD 
(ax. 1). Q. E. D. 

Again, because the two adjacent angles EGB, BGH, are 
together equal to two right angles (t"h. 6) ; of which the an- 
gle EGB has been shown to be equal to the angle GHD ; 
therefore the two angles BGH, GHD, taken together, are 
also equal to two right angles. 

Corol. 1. And, conversely, if one line meeting two other 
lines make the angles on the same side of it equal, those 
two lines are parallels. 

Corol. 2. If a line cutting two other lines make the sum 
of the two inward angles on the same side less than two 
right angles, those two lines will not be parallel, but will 
meet each other when produced. 

THEOREM XVI. 

Those lines which are parallel to the same line are parallel 
to each other. 

Let the lines AB, CD, be each of them a G B 
parallel to the line EF ; then shall the lines ~~ 71 
AB, CD, be parallel to each other. £ „l p 

For, let the line GI be perpendicular to E » _ ' I » . ■ F 
EF. Then will this line be also perpendic- I 

ular to both the lines AB, CD (corol., th. 13), and, conse- 
quently, the two lines AB, CD, are parallels (corol., th. 14). 
Q. E. D. 

THEOREM XVH. 

When one side of a triangle is produced, the exterior angle 
is equal to both the interior opposite angles taken together. 

Let the side AB, of the triangle ABC, <j jj 

be produced to D ; then will the exterior 
angle CBD be equal to the sum of the / \ / 

two interior opposite angles A and C. 

For, conceive BE to be drawn parallel 
to the side AC of the triangle. Then BC, 




A 




18 GEOMETRY. 

meeting the two parallels AC, BE, makes the alternate an- 
gles C and CBE equal (th. 13). And AD, cutting the same 
two parallels AC, BE, makes the interior and exterior an- 
gles on the same side, A and EBD, equal to each other (th. 
14). Therefore, by equal additions, the sum of the two an- 
gles A and C is equal to the sum of the two CBE and 
EBD ; that is, to the whole angle CBD (by ax. 2). Q. E. D. 

THEOREM XVIII. 

In any triangle, the sum of all the three angles is equal to 
two right angles. 

Let ABC be any plane triangle; then 
the sum of the three angles A-j-B-j-C is 
equal to two right angles. 

For, let the side AB be produced to D. 
Then the exterior angle CBD is equal to 
the sum of the two interior opposite angles A-fC (th. 17). 
To each of these equals add the interior angle B ; then will 
the sum of the three interior angles A+B+C be equal to 
the sum of the two adjacent angles ABC + CBD (ax. 2). 
But the sum of these two last adjacent angles is equal to 
two right angles (th. 6). Therefore, also, the sum of the 
three angles of the triangle A-j- B-f-C is equal to two right 
angles (ax. 1). Q. E. D. 

Corol. 1. If two angles in one triangle be equal to two 
angles in another triangle, the third angles will also be equal 
(ax. 3), and the two triangles equiangular. 

Corol. 2. If one angle in one triangle be equal to one an- 
gle in another, the sums of the remaining angles will also be 
equal (ax. 3). 

Corol. 3. If one angle of a triangle be right, the sum of 
the other two will also be equal to a right angle, and each 
of them singly will be acute, or less than a right angle. 

Corol. 4. The two least angles of every triangle are acute, 
or each less than a right angle. 



THEOREMS, 



19 




THEOREM XIX. 

In any figure whatever, the sum of all the interior angles, 
taken together, is equal to twice as many right angles, wanting 
four, as the figure has sides. 

Let ABODE be any figure ; then 
the sum of all its interior angles, A-f- 
B-f-C+D+E, is equal to twice as 
many right angles, wanting four, as 
the figure has sides. 

For, from any point F, within it, 
draw lines, FA, FB, FC, &c, to all 
the angles, dividing the polygon into 
as many triangles as it has sides, 
angles of each of these triangles is equal to two right an- 
gles (th. 18) ; therefore the sum of the angles of all the tri- 
angles is equal to twice as many right angles as the figure 
has sides. But the sum of all the angles about the point F, 
which are so many of the angles of the triangles, but no part 
of the interior angles of the polygon, is equal to four right 
angles (corol. 3, th. 6), and must be deducted out of the 
former sum. Hence it follows that the sum of all the in- 
terior angles of the polygon alone, A+B-f-C+D-f-E, is 
equal to twice as many right angles as the figure has sides, 
wanting the said four right angles. Q. E. D. 



A. B 

Now the sum of the three 



THEOREM XX. 

When every side of any figure is produced out, the sum of 
all the exterior angles thereby made is equal to four right angles. 

Let A, B, C, &c, be the exterior angles 
of any polygon, made by producing all. the 
sides; then will the sum A+B + C-j-D-f- 
E, of all those exterior angles, be equal to 
four right angles. 

For every one of these exterior angles, 
together with its adjacent interior angle, 
make up two right angles, as A-j-a equal to two right an- 
gles, being the two angles made by one line meeting another 
(th. 6). And there being as many exterior or interior an- 




20 GEOMETRY. 

gles as the figure has sides, therefore the sum of all the in- 
terior and exterior angles is equal to twice as many right 
angles as the figure has sides. But the sum of all the inte- 
rior angles, with four right angles, is equal to twice as many 
right angles as the figure has sides (th. 19) ; therefore the 
sum of all the interior and all the exterior angles is equal 
to the sum of all the interior angles and four right angles 
(by ax. 1). From each of these take away all the interior 
angles, and there remain all the exterior angles equal to four 
right angles (by ax. 3). Q. E. D. 

THEOREM XXI. 

A 'perpendicular is the shortest line that can he drawn from 
a given point to an indefinite line. And, of any other lines 
drawn from the same point, those that are nearest the perpen- 
dicular are less than those more remote. 

If AB, AC, AD, &c, be lines drawn from 
the given point A to the indefinite line DE, 
of which AB is perpendicular, then shall the 
perpendicular AB be less than AC, and AC 
less than AD, &c. 

For, the angle B being a right one, the an- ^ C B E 
gle C is acute (by cor. 3, th. 18), and therefore less than the 
angle B. But the less angle of a triangle is subtended by 
the less side (th. 9). Therefore the side AB is less than the 
side AC 

Again, the angle ACB being acute, as before, the adjacent 
angle ACD will be obtuse (by th. 6) ; consequently, the an- 
gle D is acute (corol. 3, th. 18), and therefore is less than 
the angle C. And since the less side is opposite to the 
less angle, therefore the side AC is less than the side AD. 
Q. E. D. 

Corol. A perpendicular is the least distance of a given 
point from a line. 




THEOREMS. 21 



THEOREM XXH. 




The opposite sides and opposite angles of any parallelogram 
are equal to each other ; and the diagonal divides it into two 
equal triangles. 

Let ABCD be a parallelogram, of which 
the diagonal is BD ; then will its opposite 
sides and opposite, angles be equal to each 
other, and the diagonal BD will divide it 
into two equal parts, or triangles. 

For, since the sides AB and DC are parallel, as also the 
sides AD and BC (defin. 32), and the line BD meets them, 
therefore the alternate angles are equal (th. 13), namely, the 
angle ABD to the angle CDB, and the angle ADB to the 
angle CBD. Hence the two triangles, having two angles 
in the one equal to two angles in the other, have also their 
third angles equal (cor. 1, th. 18), namely, the angle A equal 
to the angle C, which are two of the opposite angles of the 
parallelogram. 

Also, if to the equal angles ABD, CDB, be added the 
equal angles CBD, ADB, the wholes will be equal (ax. 2), 
namely, the whole angle ABC to the whole ADC, which 
are the other two opposite angles of the parallelogram. 
Q. E. D. 

Again, since the two triangles are mutually equiangular, 
and have a side in each equal, viz., the common side BD, 
therefore the two triangles are identical (th. 2), or equal in 
all respects, namely, the side AB equal to the opposite side 
DC, and AD equal to the opposite side BC, and the whole 
triangle ABD equal to the whole triangle BCD. Q. E. D. 

Corol. 1. Hence, if one angle of a parallelogram be a right 
angle, all the other three will also be right angles, and the 
parallelogram a rectangle. 

Corol. 2- Hence, also, the sum of any two adjacent angles 
of a parallelogram is equal to two right angles. 



22 GEOMETRY. 



THEOREM XXm, 




Every quadrilateral ivhose opposite sides are equal is a par- 
allelogram, or has the opposite sides parallel. 

Let ABCD be a quadrangle, having the 
opposite sides equal, namely, the side AB 
equal to DC, and AD equal to BC ; then 
shall these equal sides be also parallel, and 
the figure a parallelogram. 

For, let the diagonal BD be drawn. Then, the triangles 
ABD, CBD, being mutually equilateral (by hyp.), they are 
also mutually equiangular (th. 5), or have their correspond- 
ing angles equal ; consequently, the opposite sides are par- 
allel (th. 14); viz., the side AB parallel to DC, and AD 
parallel to BC, and the figure is a parallelogram. Q. E. D. 

THEOREM XXIV. 

Those lines which join the corresponding extremes of two equal 
and parallel lines are themselves equal and parallel. 

Let AB, DC, be two equal and parallel lines ; then will 
the lines AD, BC, which join their extremes, be also equal 
and parallel. [See the fig. above.] 

For, draw the diagonal BD. Then, because AB and 
DC are parallel (by hyp.), the angle ABD is equal to the 
alternate angle BDC (th. 13). Hence, then, the two trian- 
gles, having two sides and the contained angles equal, viz., 
the side AB equal to the side DC, and the side BD com- 
mon, and the contained angle ABD equal to the contained 
angle BDC, they have the remaining sides and angles also 
respectively equal (th. 1) ; consequently, AD is equal to BC, 
and also parallel to it (th. 13). Q. E. D. 

THEOREM XXV. 

Parallelograms, as also triangles, standing on the same base 
and between the same parallels, are equal to each other. 

Let ABCD, ABEF, be two parallelograms, and ABC, 
ABF, two triangles, standing on the same base, AB, and be- 
tween the same parallels AB, DE ; then will the parallelo- 



THEOREMS, 



23 




gram ABCD be equal to the parallelo- jy 
gram ABEF, and the triangle ABC equal 
to the triangle ABF. 

For, since the line DE cuts the two par- 
allels AF, BE, and the two AD, BC, it 
makes the angle E equal to the angle 
AFD, and the angle D equal to the angle BCE (th. 15) ; 
the two triangles ADF, BCE, are therefore equiangular 
(cor. 1, th. 18) ; and having the two corresponding sides 
AD, BC, equal (th. 23), being opposite sides of a parallelo- 
gram, these two triangles are identical, or equal in all re- 
spects (th. 2). If each of these equal triangles, then, be taken 
from the whole space ABED, there will remain the paral- 
lelogram ABEF in the one case, equal to the parallelogram 
ABCD in the other (by ax. 3). 

Also, the triangles AJ3C, ABF, on the same base AB, and 
between the same parallels, are equal, being the halves of 
the said equal parallelograms (th. 23). Q. E. D. 

Corol 1. Parallelograms or triangles having the same 
base and altitude are equal. For the altitude is the same 
as the perpendicular or distance between the two parallels, 
which is every where equal, by the definition of parallels. 

Corol 2. Parallelograms or triangles having equal bases 
and altitudes are equal. For, if the one figure be applied 
with its base on the other, the bases will coincide or be the 
same, because they are equal ; and so the two figures, hav-' 
ing the same base and altitude, are equal. 



THEOREM XXVI. 

If a parallelogram and a triangle stand on the same base 
and between the same parallels, the parallelogram will be double 
the triangle, or the triangle half the parallelogram. 

Let ABCD be a parallelogram, and ABE D C 
a triangle, on the same base AB, and between 
the same parallels AB, DE; then will the 
parallelogram ABCD be double the triangle 
ABE, or the triangle half the parallelogram. 

For, draw the diagonal AC of the parallel- 
ogram, dividing it into two equal parts (th. 
22) 




A B 

Then, because the triangles ABC, ABE, on the same 



24 GEOMETRY. 

base and between the same parallels are equal (th. 25), and 
because the one triangle ABC is half the parallelogram 
ABCD (th. 23), the other equal triangle ABE is also equal 
to half the same parallelogram ABCD. Q. E. D. 

Corol. 1. A triangle is equal to half a parallelogram of 
the same base and altitude, because the altitude is the per- 
pendicular distance between the parallels, which is every- 
where equal, by the definition of parallels. 

Corol. 2. If the base of a parallelogram be half that of a 
triangle of the same altitude, or the base of the triangle be 
double that of the parallelogram, the two figures will be 
equal to each other. 

THEOREM XXVII. 

Rectangles that are contained by equal lines are equal to 
each other. 

Let BD, FH, be two rectangles, hav- 
ing the sides AB, BC, equal to the sides 
EF, FGr, each to each ; then will the 
rectangle BD be equal to the rectan- 
gle FH. 

For, draw the two diagonals AC, 
EG-, dividing the two parallelograms 
each into two equal parts. Then the two triangles ABC, 
EFG-, are equal to each other (th. 1), because they have the 
two sides AB, BC, and the contained angle B, equal to the 
two sides EF, FG, and the contained angle F (by hyp.). 
But these equal triangles are the halves of the respective 
rectangles. And because the halves or the triangles are 
equal, the wholes, or the rectangles DB, HF, are also equal 
(by ax. 6). Q. E. D. 

Corol. The squares on equal lines are also equal; for 
every square is a species of rectangle. 

THEOREM XXVIII. 

The complements of the parallelograms which are about the 
diagonal of any parallelogram are equal to each other. 

Let AC be a parallelogram, BD a diagonal, EIF parallel 
to AB or DC, and GIH parallel to AD or BC, making AT, 




D G 


C 


E f\^ 


lv 


1'N 



A H 



THEOREMS. 25 

IC, complements to the parallelograms EG, 
HF, which are about the diagonal DB ; 
then will the complement AI be equal to 
the complement IC. 

For, since the diagonal DB bisects the 
three parallelograms AC, EG, HF (th. 23), 
therefore, the whole triangle DAB being equal to the whole 
triangle DCB, and the parts DEI, IHB, respectively equal 
to the parts DGI, IFB, the remaining parts AI, IC, must 
also be equal (by ax. 3). Q. E. D. 

THEOREM XXIX. 

A trapezoid or quadrilateral having two sides parallel is equal 
to half a parallelogram whose base is the sum of these two sides, 
and its altitude the perpendicular distance between them. 

Let ABCD be the trapezoid, having its two ^ ^ 
sides AB, DC, parallel ; and in AB produced 
take BE equal to DC, so that AE may be the 
sum of the two parallel sides ; produce DC 
also, and let EF, GC, BH, be all three par- 
allel to AD. Then is AF a parallelogram of 
the same altitude with the trapezoid ABCD, having its base 
AE equal to the sum of the parallel sides of the trapezoid ; 
and it is to be proved that the trapezoid ABCD is equal to 
half the parallelogram AF. 

Now, since triangles or parallelograms of equal bases 
and altitude are equal (corol. 2, th. 25), the parallelogram 
DG is equal to the parallelogram HE, and the triangle CGB 
equal to the triangle CHB ; consequently, the line BC bi- 
sects, or equally divides, the parallelogram AF, and ABCD 
is the half of it. Q. E. D. 

THEOREM XXX. 

The sum of all the rectangles contained under one ivhole line, 
and the several parts of another line any way divided, is equal 
to the rectangle contained under the two ivhole lines. 

Let AD be the one line, and AB the other, divided into 
the parts AE, EF, FB ; then will the rectangle contained 
by AD and AB be equal to the sum of the rectangles of 

B 




26 GEOMETRY. 

AD and AE, and AD and EF, and AD jy Gl- 
and FB ; thus expressed, AD . AB — 
AD . AE+AD . EF+AD . FB. 

For, make the rectangle AC of the 
two whole lines AD, AB, and draw ^ 
EG, FH, perpendicular to AB, or parallel to AD, to which 
they are equal (th. 24). Then the whole rectangle AC is 
made up of all the other rectangles AG, EH, FC. But 
these rectangles are contained by AD and AE, EG and 
EF, FH and FB ; which are equal to the rectangles of 
AD and AE, AD and EF, AD and FB, because AD is 
equal to each of the two, EG, FH. Therefore the rectan- 
gle AD . AB is equal to the sum of all the other rectangles 
AD . AE, AD . EF, AD . FB. Q. E. D. 

Corol. If a right line be divided into any two parts, the 
square on the whole line is equal to both the rectangles of 
the whole line and each of the parts. 

THEOREM XXXI. 

The square of the sum of two lines is greater than the sum 
of their squares oy twice the rectangle of the said lines. Or, 
the square of a whole line is equal to the sum of the squares of 
its two parts, together with twice the rectangle of those parts. 

Let the line AB be the sum of any two lines 
AC, CB ; then will the square of AB be equal 
to the squares of AC, CB, together with twice 
the rectangle of AC . CB. That is, AB 2 =: 
AC 2 +CB 2 +2AC . CB. A 1 — — T!b 

For, let ABDE be the square on the sum or whole line 
AB, and ACFG the square on the part AC. Produce CF 
and GF to the other sides at H and I. 

From the lines CH, GI, which are equal, being each equal 
to the sides of the square AB or BD (th. 24), take the parts 
CF, GF, which are also equal, being the sides of the square 
AF, and there remains FH equal to FI, which are also 
equal to DH, DI, being the opposite sides of the parallelo- 
gram. Hence the figure HI is equilateral; and it has all 
its angles right ones (corol. 1, th. 22) ; it is therefore a 
square on the line FI, or the square of its equal CB. Also 
the figures EF. FB, are equal to two rectangles under AC 



THEOREMS. 27 

and CB, because GF is equal to AC,' and FH or FI equal 
to CB. But the whole square AD is made up of the four 
figures, viz., the two squares AF, FD, and the two equal 
rectangles EF, FB. That is, the square of AB is equal to 
the squares of AC, CB, together with twice the rectangle 
of AC, CB. Q. E. D. 

Corol. Hence, if a line be divided into two equal parts, 
the square of the whole line will be equal to four times the 
square of half the line. 

THEOREM XXXH. 

The square of the difference of two lines is less than the sum 
of their squares by twice the rectangle of the said lines. 



Let AC, BC, be any two lines, and AB their ^ 
difference ; then will the square of AB be less E 
than the squares of AC, BC, by twice the rect- 
angle of AC and BC. Or, AB 2 =AC 2 +BC 2 
-2AC.BC. A 



For, let ABDE be the square on the differ- 



ence AB, and ACFG- the square on the line AC. Produce 
ED to H ; also produce DB and HC, and draw KI, making 
BI the square of the other line BC. 

Now it is visible that the square AD is less than the two 
squares AF, BI, by the two rectangles EF, DI. But GF 
is equal to the one line AC, and GE or FH is equal to the 
other line BC ; consequently the rectangle EF, contained 
under EG and GF, is equal to the rectangle of AC and BC. 

Again : FH being equal to CI or BC or DH, by adding 
the common part HC, the whole HI will be equal to the 
whole FC, or equal to AC ; and, consequently, the figure 
DI is equal to the rectangle contained by AC and BC 

Hence the two figures EF, DI, are two rectangles of the 
two lines AC, BC ; and, consequently, the square of AB is 
less than the squares of AC, BC, by twice the rectangle 
AC.BC. Q. E. D. 



28 GEOMETRY. 



THEOEEM XXXIII. 



The rectangle under the sum and difference of two lines is 
squal to the difference of the squares of those lines.* 

Let AB, AC, be any two unequal lines ; E ^ 3 

then will the difference of the squares of AB, ^ ~ 
AC, be equal to a rectangle under their sum 
and difference. That is, 



AB 2 -AC 2 =AB+AC . AB-AC. 

For, let ABDE be the square of AB, and 
ACFG the square of AC. Produce DB till 
BH be equal to AC ; draw HI parallel to AB or ED, and 
produce FC both ways to I and K. 

Then the difference of the two squares AD, AF, is evi- 
dently the two rectangles EF, KB. But the rectangles EF, 
BI, are equal, being contained under equal lines ; for EK 
and BH are each equal to AC, and GE is equal to CB, be- 
ing each equal to the difference between AB and AC, or 
their equals AE and AG. Therefore the two EF, KB, are 
equal to the two KB, BI, or to the whole KH ; and, conse- 
quently, KH is equal to the difference of the squares AD, 
AF. But KH is a rectangle contained by DH, or the sum of 
AB and AC, and by KD, or the difference of AB and AC ; 
therefore the difference of the squares of AB, AC, is equal 
to the rectangle under their sum and difference. Q. E. D. 

THEOREM XXXIV. 

In any right-angled triangle, the square of the hypothenuse is 
equal to the sum of the squares of the other two sides. 

Let ABC be a right-angled triangle, having the right an- 
gle A ; then will the square of the hypothenuse CB be equal 
to the sum of the squares of the other two sides AC, AB. 
Or CB 2 =AC 2 +AB 2 . 

For, on CB describe the square CE, and on AC, AB, the 

* This and the two preceding theorems are evinced algebraically 
by the three expressions, 

(a + 6) 2 :=a 2 +2a&+& 2 =a 2 +& 2 +2aZ>; 
(a-by=a 2 -2ab + b 2 =a-\-b 2 -2ab; 
(a + 6) (a-b)=a 2 -b 2 . 



THEOREMS. 



29 




squares CG, BH ; then draw AK 
parallel to CD or BE ; and join 
CI, BF, AD, AE. 

Now, because the line AC meets 
the two AG-, AB, so as to make 
two right angles, these two form 
one straight line GB (corol. 1, th. 
6). And because the angle FCA 
is equal to the angle DCB, being 
each a right angle, or the angle of 
a square ; to each of these equals D L E 

add the common angle BCA, so will the whole angle or sum 
FCB be equal to the whole angle or sum ACD. But the 
line FC is equal to the line AC, and the line CB to the line 
CD, being sides of the same squares ; so that the two sides 
FC, CB, and their included angle FCB, are equal to the two 
sides CA, CD, and the contained angle ACD, each to each ; 
therefore the whole triangle CFB is equal to the whole tri- 
angle ACD (th. 1). 

But the square CG- is double the triangle CFB, on the 
same base FC, and between the same parallels FC, GB (th. 
26) ; in like manner, the parallelogram CK is double the 
triangle ACD, on the same base CD, and between the same 
parallels CD, AK. And since the doubles of equal things 
are equal (by ax. 6), therefore the square CG is equal to 
the parallelogram CL. 

In like manner, the other square BH is proved equal to 
the other parallelogram BL ; consequently, the two squares 
CG and BH together are equal to the two parallelograms 
CL and BL together, or to the whole square CE. That is, 
the sum of the two squares on the two less sides is equal to 
the square on the greatest side. Q. E. D. 

Corol. 1. Hence the square of either of the two less sides 
is equal to the difference of the squares of the hypothenuse 
and the other side (ax. 3), or equal to the rectangle con- 
tained by the sum and difference of the said hypothenuse 
and other side (th. 33). 

Corol. 2. Hence, also, if two right-angled triangles have 
two sides of the one equal to two corresponding sides of the 
other, their third sides will also be equal, and the triangles 
identical. 



GEOMETRY. 



THEOREM XXXV. 



In any triangle, the difference of the squares of the tivo sides 
is equal to the difference of the squares of the segments of the 
base, or of the two lines or distances included between the ex- 
tremes of the base and the perpendicular. 

Let ABC be any triangle, having C C 

CD perpendicular to AB ; then will a A 

the difference of the squares of AC, /\ / l\ 

BC, be equal to the difference of the / I \ / \ \ 

squares of AD, BD ; that is, AC 2 — ^~ i-1 Z L_\ 

BC 2 =AD 2 — BD 2 . A BDA D B 

For, since AC 2 is equal to AD 2 +CD 2 ") „ , 
and BC 2 is equal to BD 2 +CD 2 | W th ' 6 ^> 

therefore the difference between AC 2 and BC 2 is equal to 
the difference between AD 2 -fCD 2 and BD 2 +CD 2 , or equal 
to the difference between AD 2 and BD 2 , by taking away the 
common square CD 2 . Q. E. D. 

Corol. The rectangle of the sum and difference of the two 
sides of any triangle is equal to the rectangle of the sum 
and difference of the distances between the perpendicular 
and the two extremes of the base, or equal to the rectangle 
of the base and the difference or sum of the segments, ac- 
cording as the perpendicular falls within or without the tri- 
angle. That is, 

(AC+BC) . (AC-BC)^(AD-(-BD) . (AD-BD). 
Or (AC+BC) . (AC-BC)=AB (AD-BD) in the 2d fig., 
and (AC+BC). (AC-BC)=AB.(AD-j-BD) in the 1st fig. 

THEOREM XXXVI. 

In any obtuse-angled triangle, the square of the side subtend- 
ing the obtuse angle is greater than the sum of the squares of 
the other two sides by twice the rectangle of the base and the 
distance of the perpendicular from the obtuse angle. 

Let ABC be a triangle obtuse angled at B, and CD per- 
pendicular to AB ; then will the square of AC be greater 
than the squares of AB, BC, by twice the rectangle of AB, 
BD. That is, AC 2 = AB 2 +BC 2 +2AB . BD. See the 1st 
figure above. 



THEOREMS. 



31 



For AD 2 =AB 2 +BD 2 -h2AB . BD (th. 31). 
And AD 2 +CD 2 =AB 2 -|-BD 2 H-CD 2 4-2AB . BD (ax. 2). 
But AD 2 +CD 2 =AC 2 , and BD 2 +CD 2 =BC 2 (th. 34). 
Therefore AC 2 =AB 2 +BC 2 +2AB . BD. Q. E. D. 



THEOREM XXXVn. 

In any triangle, the square of the side subtending an acute 
angle is less than the squares of the base and the other side by 
twice the rectangle of the base and the distance of the perpen- 
dicular from the acute angle. 

Let ABC be a triangle, having the 
angle A acute, and CD perpendicu- 
lar to AB ; then will the square of 
BC be less than the squares of AB, 

AC, by twice the rectangle of AB, 

AD. That is, BC 2 = AB 2 -fAC 2 - 
2AD.AB. 

For BD 2 = AD 2 -f AB 2 -2AD . AB (th. 32). 

And BD 2 +DC 2 =AD 2 +DC 2 +AB 2 -2AD. AB (ax. 2); 
therefore BC 2 =AC 2 + AB 2 -2AD . AB (th. 34). Q. E. D. 




B D A 



D B 



THEOREM XXXVTH. 

In any triangle, the double of the square of a line draivn 
from the vertex to the 'middle of the base, together with double 
the square of the half base, is equal to the sum of the squares 
of the other two sides. 

Let ABC be a triangle, and CD the line 
drawn from the vertex to the middle of the 
base AB, bisecting it into the two equal 
parts AD, DB ; then will the sum of the 
squares of AC, CB, be equal to twice the 
sum of the squares of CD, AD ; or AC 2 -f- 
CB 2 =2CD 2 +2AD 2 . 

For AC 2 =zCD 2 + AD 2 +2AD . DE (th. 36). 

And BC 2 =CD 2 -f BD 2 -2AD . DE (th. 37). 

Therefore AC 2 +BC 2 z=:2CD 2 +AD 2 +BD 2 . 

=2CD 2 +2AD 2 (ax. 2). Q. E. D. 





32 GEOMETRY. 



THEOREM XXXIX. 

In an isosceles triangle, the square of a line drawn from the, 
vertex to any point in the base, together with the rectangle of the 
segments of the base, is equal to the square of one of the equal 
sides of the triangle. 

Let ABC be the isosceles triangle, and 
CD a line drawn from the vertex to any 
point D in the base; then will the square 
of AC be equal to the square of CD, togeth- 
er with the rectangle of AD and DB. That X' d 
is,AC 2 =CD 2 +AD.DB. 

For AC 2 -CD 2 =AE 2 -DE 2 (th. 35). 

= AD . DB (th. 33). 

Therefore AC 2 =CD 2 + AD . DB (ax. 2). Q. E. D. 

THEOREM XL. 

In any parallelogram the two diagonals bisect each other; 
and the sum of their squares is equal to the sum of the squares 
of all the four sides of the parallelogram. 

Let ABCD be a parallelogram whose ^ D 

diagonals intersect each other in E ; then 
will AE be equal to EC, and BE to ED ; 
and the sum of the squares of AC, BD, 
will be equal to the sum of the squares 
of AB, BC, CD, DA. That is, 

AE=EC, and BE=ED, 
and AC 2 -fBD 2 =AB 2 -hBC 2 +CD 2 -| : DA 2 . 

For the triangles AEB, DEC, are equiangular, because 
they have the opposite angles at E equal (th. 7), and the 
two lines AC, BD, meeting the parallels AB, DC, make the 
angle BAE equal to the angle DCE, and the angle ABE 
equal to the angle CDE, and the side AB equal to the side 
DC (th. 22) ; therefore these two triangles are identical, and 
have their corresponding sides equal (th. 2), viz., AE=EC, 
and BE=ED. 

Again, since AC is bisected in E, the sum of the squares 
AD 2 +DC 2 =2AE 2 +2DE 2 (th. 38). 

In like manner, AB 2 +BC 2 =2AE 2 +2BE 2 or 2DE 2 . 




THEOREMS. 33 

Therefore AB 2 +BC 2 -hCD 2 +DA 2 ==4AE 2 +4DE 2 (ax.2). 

But because the square of a whole line is equal to four 
times the square of half the line (cor., th. 31), that is, AC 2 = 
4AE 2 ,andBD 2 =4DE 2 ; 

Therefore AB 2 +BC 2 +CD 2 +DA 2 =AC 2 +BD 2 (ax. 1). 
Q. E. D. 

Cor. 1. If AD = DC, or the parallelogram be a rhombus, 
then AD 2 =AE 2 +ED 2 , CD 2 =DE 2 +CE 2 , &c. 

Cor. 2. Hence, and by th. 34, the diagonals of a rhombus 
intersect at right angles. 

B2 






■ 




BOOK II. 
OF THE CIECLE. 

THEOEEM I. 

If a line, drawn through or from the centre of a circle, bi- 
sect a chord, it will be perpendicular to it ; or, if it be perpen- 
dicular to the chord, it will bisect both the chord and the arc of 
the chord. 

Let AB be any chord in a circle, and CD 
a line drawn from the centre C to the chord. 
Then, if the chord be bisected in the point D, 
CD will be perpendicular to AB. 

Draw the two radii CA, CB. Then the 
two triangles ACD, BCD, having CA equal 
to CB (def. 44), and CD common, also AD equal to DB 
(by hyp.) ; they have all the three sides of the one equal to 
all the three sides of the other, and so have their angles also 
equal (th. 5, B. I.). Hence, then, the angle ADC being 
equal to the angle BDC, these angles are right angles, and 
the line CD is perpendicular to AB (def. 11). 

Again, if CD be perpendicular to AB, then will the 
chord AB be bisected at the point D, or have AD equal to 
DB ; and the arc AEB bisected in the point E, or have 
AE equal EB. 

For, having drawn CA, CB, as before ; then, in the tri- 
angle ABC, because the side CA is equal to the side CB, 
their opposite angles A and B are also equal (th. 3). Hence, 
then, in the two triangles ACD, BCD, the angle A is equal 
to the angle B, and the angles at D are equal (def. 11); 
therefore their third angles are also equal (corol. 1, th. 17, 
B. I.). And having the side CD common, they have also 
the side AD equal to the side DB (th. 2, B. I.). 

Also, since the angle ACE is equal to the angle BCE, 
the arc AE, which measures the former (def. 57), is equal 
to the arc BE, which measures the latter, since equal angles 
must have equal measures. 



THEOREMS. 



35 



Corol. Hence a line bisecting any chord at right angles 
passes through the centre of the circle. 




THEOREM H. 

If more than two equal lines can he drawn from any point 
within a circle to the circumference, that point will be the centre. 

Let ABC be a circle, and D a point 
within it ; then, if any three lines, DA, 
DB, DC, drawn from the point D to the 
circumference, be equal to each other, the 
point D will be the centre. 

Draw the chords AB, BC, which let be 
bisected in the points E, G, and join DE, DG. 

Then the two triangles DAE, DBE, have the side DA 
equal to the side DB by supposition, and the side AE equal 
to the side EB by hypothesis, also the side DE common; 
therefore these two triangles are identical, and have the an- 
gles at E equal to each other (th. 5) ; consequently, DE is 
perpendicular to the middle of the chord AB (def. 11), and 
therefore passes through the centre of the circle (corol., th. 1). 

In like manner, it may be shown that DG passes through 
the centre. Consequently, the point D is the centre of the 
circle, and the three equal lines DA, DB, DC, are radii. 
Q. E. D. 



THEOREM HI. 

If tiuo circles, placed one within another, touch, the centres 
of the circles and the point of contact will be all in the same 
right line. 

Let the two circles ABC, ADE, touch one 
another internally in the point A ; then will 
the point A and the centres of those circles 
be all in the same right line. 

Let F be the centre of the circle ABC, 
through which draw the diameter AFC. 
Then, if the centre of the other circle can be 
out of this line AC, let it be supposed in some other point, 
as G, through which draw the line FG, cutting the two cir- 
cles in B and D. 




36 GEOMETRY. 

Now, in the triangle AFG, the sum of the two sides FG, 
GA, is greater than the side AF (th. 10, B. L), or greater 
than its equal radius FB. From each of these take away the 
common part FG, and the remainder GA will be greater 
than the remainder GB. But the point G being supposed 
the centre of the inner circle, its two radii, GA, GD, are 
equal to each other ; consequently, GD will also be greater 
than GB. But ADE being the inner circle, GD is neces- 
sarily less than GB. So that GD is both greater and less 
than GB, which is absurd. To get rid of this absurdity, 
we must abandon the supposition that produced it, which 
was that G might be out of the line AFC. Consequently, 
the centre G can not be out of the line AFC. Q. E. D. 

THEOREM IV. 

If two circles touch one another externally, the centres of the 
circles and the point of contact will be all in the same right line. 

Let the two circles ABC, ADE, touch one 
another externally at the point A ; then will 
the point of contact A and the centres of the 
two circles be all in the same right line. 

Let F be the centre of the circle ABC, 
through which draw the diameter AFC, and 
produce it to the other circle at E. Then, 
if the centre of the other circle ADE can be 
out of the line FE, let it, if possible, be sup- 
posed in some other point, as G, and draw 
the lines AG, FBDG, cutting the two circles 
in B and D. 

Then, in the triangle AFG, the sum of the two sides AF, 
AG, is greater than the third side FG (th. 10, B. I.). But 
F and G being the centres of the two circles, the two radii 
GA, GD, are equal, as are also the two radii AF, FB. 
Hence the sum of GA, AF, is equal to the sum of GD, BF ; 
and therefore this latter sum also, GD, BF, is greater than 
GF, which is absurd. Consequently, the centre G can not 
be out of the line EF. Q. E. D. 




THEOREMS. 



THEOREM V. 




Any chords in a circle ivhich are equally distant from the 
centre are equal to each other ; or, if they be equal to each other, 
they ivill be equally distant from the centre. 

Let AB, CD, be any two chords at equal 
distances from the centre, G ; then will these 
two chords, AB, CD, be equal to each other. 

Draw the two radii G-A, GC, and the two 
perpendiculars GE, GF, which are the equal 
distances from the centre, G. Then, the two 
right-angled triangles, GAE, GCF, having the side GA 
equal the side GC, and the side GE equal the side GF, and 
the angle at E equal to the angle at F, therefore those two 
triangles are identical (Cor. 2, th. 34, B. I.), and have the 
line AE equal to the line CF. But AB is the double of 
AE, and CD is the double of CF (th. 1) ; therefore AB is 
equal to CD (by ax. 6). Q. E. D. 

Again, if the chord AB be equal to the chord CD, then 
will their distances from the centre, GE, GF, also be equal 
to each other. 

For, since AB is equal CD by supposition, the half AE 
is equal the half CF. Also, the radii GA, GC, being equal, 
as well as the right angles E and F, therefore the third sides 
are equal (Cor. 2, th. 34), or the distance GE equal the dis- 
tance GF. Q. E. D. 

THEOREM VI. 

A line perpendicular to a radius at its extremity is a tan- 
gent to the circle. 

Let the line ADB be perpendicular to the 
radius CD of a circle ; then shall AB touch 
the circle in the point D only. 

From any other point, E, in the line AB, 
draw CFB to the centre, cutting the circle 
in F. 

Then, because the angle D of the triangle CDE is a right 
angle, the angle at E is acute (Cor. 3, th. 17, B. I.), and, 
consequently, less than the angle D. But the greater side 




38 GEOMETRY. 

is always opposite to the greater angle (tli. 9, B. I.), there- 
fore the side CE is greater than the side CD, or greater than 
its equal CF. Hence the point E is without the circle; 
and the same for every other point in the line AB. Con- 
sequently the whole line is without the circle, and meets it 
in the point D only. 

THEOREM VII. 

When a line is a tangent to a circle, a radius drawn to the 
point of contact is perpendicular to the tangent. 

Let the line AB touch the circumference of a circle at the 
point D ; then will the radius CD be perpendicular to the 
tangent AB. [See the last figure.] 

For, the line AB being wholly without the circumference 
except at the point D, every other line, as CE, drawn from 
the centre C to the line AB, must pass out of the circle to 
arrive at this line. The line CD is therefore the shortest 
that can be drawn from the point C to the line AB, and, 
consequently (th. 21, B. I.), it is perpendicular to that line. 

Corel. Hence, conversely, a line drawn perpendicular to 
a tangent at the point of contact passes through the centre 
of the circle. 

THEOREM VHI. 

The angle formed by a tangent and chord is measured by 
half the arc of that chord. 

Let AB be a tangent to a circle, and CD a chord drawn 
from the point of contact, C ; then is the angle BCD meas- 
ured by half the arc CFD, and the angle ACD measured by 
half the arc CGD. 

Draw the radius EC to the point of con- 
tact, and the radius EF perpendicular to the 
chord at H. 

Then the radius EF, being perpendicular 
to the chord CD, bisects the arc CFD (th. 
1). Therefore CF is half the arc CFD. 

In the triangle CEH, the angle H being a right one, the 
sum of the two remaining angles E and C is equal to a right 
angle (Cor. 3, th. 17, B. I.), which is equal to the angle 




THEOREMS. 39 

BCE, because the radius CE is perpendicular to the tangent. 
From each of these equals take away the common part or 
angle C, and there remains the angle E equal to the angle 
BCD. But the angle E is measured by the arc CF (def. 
57), which is the half of CFD; therefore the equal angle 
BCD must also have the same measure, namely, half the 
arc CFD of the chord CD. 

Again, the line GEF, being perpendicular to the chord 
CD, bisects the arc CGD (th. 1). Therefore CG is half 
the arc CGD. Now, since the line CE, meeting FG, makes 
the sum of the two angles at E equal to two right angles 
(th. 6, B. I.), and the line CD makes with AB the sum of 
the two angles at C equal to two right angles ; if from these 
two equal sums there be taken away the parts or angles 
CEH and BCH, which have been proved equal, there re- 
mains the angle CEG equal to the angle ACH. But the 
former of these, CEG, being an angle at the centre, is meas- 
ured by the arc CG (def. 57) ; consequently, the equal angle 
ACD must also have the same measure CG, which is half 
the arc CGD of the chord CD. Q. E. D. 

Corol. 1. The sum of the two right angles is measured by 
half the circumference. For the two angles BCD, ACD, 
which make up two right angles, are measured by the arcs 
CF, CG, which make up half the circumference, FG being 
a diameter. 

Corol. 2. Hence, also, one right angle must have for its 
measure a quarter of the circumference, or 90 degrees. 

THEOREM IX. 

An angle at the circumference of a circle is measured by half 
the arc that subtends it. 

Let BAC be an angle at the circumfer- DAE 
ence : it has for its measure half the arc T/^TX 
BC, which subtends it. / / \ \ 

For, suppose the tangent DE to pass [/ \ I 
through the point of contact A ; then, the B\^ ^/Q 
angle DAC being measured by half the 
arc ABC, and the angle DAB by half the arc AB (th. 8), 
it follows, by equal subtraction, that the difference, or angle 
BAC, must be measured by half the arc BC, which it stands 
upon. Q. E. D. 



40 



GEOMETRY. 



THEOREM X. 

All angles in the same segment of a circle, or standing on 
same arc, are equal to each other. 

Let C and D be two angles in the same 
segment ACDB, or, which is the same thing, 
standing on the supplemental arc AEB ; 
then will the angle C be equal to the angle D. 

For each of these angles is measured by 
half the arc AEB ; and thus, having equal 
measures, they are equal to each other 
(ax. 11). 




THEOREM XI. 

An angle at the centre of a circle is double the angle at the 
circumference, when both stand on the same arc. 

Let C be an angle at the centre C, and 
D an angle at the circumference, both 
standing on the same arc or same chord 
AB ; then will the angle C be double of 
the angle D, or the angle D equal to half 
the angle C. 

For the angle at the centre C is meas- 
ured by the whole arc AEB (def. 57), and the angle at the 
circumference D is measured by half the same arc AEB (th. 
9) ; therefore the angle D is only half the angle C, or the 
angle C double the angle D. 




THEOREM XII. 

An angle in a semicircle is a right angle. 

If ABC or ADC be a semicircle, then any angle D in 
that semicircle is a right angle. 

For the angle D, at the circumference, is measured by 
half the arc ABC (th. 9), that is, by a quadrant of the cir- 
cumference. But a quadrant is the measure of a right angle 
(Cor. 4, th. 6 ; or Cor. 2, th. 8). Therefore the angle D is 
a right angle. 



THEOREMS. 41 



THEOREM XIH. 




The angle formedby a tangent to a circle, and a chord drawn 
from the point of contact, is equal to the angle in the alternate 
segment. 

If AB be a tangent, and AC a chord, 
and D any angle in the alternate segment 
ADC, then will the angle D be equal to 
the angle BAC made by the tangent and 
chord of the arc AEC. 

For the angle D, at the circumference, 
is measured by half the arc AEC (th. 9) ; and the angle 
BAC, made by the tangent and chord, is also measured by 
the same half arc AEC (th. 8) ; therefore these two angles 
are equal (ax, 11). 

THEOREM XIV. 

The sum of any two opposite angles of a quadrangle in- 
scribed in a circle is equal to two right angles. 

Let ABCD be any quadrilateral inscribed 
in a circle ; then shall the sum of the two 
opposite angles A and C, or B and D, be 
equal to two right angles. 

For the angle A is measured by half the 
arc DCB, which it stands upon, and the an- 
gle C by half the arc DAB (th. 9) ; therefore the sum of 
the two angles A and C is measured by half the sum of 
these two arcs, that is, by half the circumference. But half 
the circumference is the measure of two right angles (Cor. 
4, th. 6, B. I.) ; therefore the sum of the two opposite angles 
A and C is equal to two right angles. In like manner it is 
shown that the sum of the other two opposite angles, D and 
B, is equal to two right angles. Q. E. D 

THEOREM XV. 

If any side of a quadrangle inscribed in a circle be produced 
out, the exterior angle will be equal to the interior opposite angle. 

If the side AB, of the quadrilateral ABCD, inscribed in 




42 



GEOMETRY. 




a circle, be produced to E, the exterior 
angle DAE will be equal to the interior 
opposite angle C. 

For the sum of the two adjacent an- 
gles DAE and DAB is equal to two right 
angles (th. 6), and the sum of the two 
opposite angles, C and DAB is also equal to two right an- 
gles (th. 14) ; therefore the former sum of the two angles 
DAE and DAB is equal to the latter sum of the two C 
and DAB (ax. 1). From each of these equals taking away 
the common angle DAB, there remains the angle DAE 
equal the angle C. Q. E. D. 



THEOREM XVI. 

Any tivo parallel chords intercept equal arcs. 

Let the two chords AB, CD, be parallel ; 
then will the arcs AC, BD, be equal ; or AC 
=DB. 

Draw the line BC Then, because the lines 
AB, CD, are parallel, the alternate angles B 
and C are equal (th. 13, B. I.). But the angle at the circum- 
ference B is measured by half the arc AC (th. 9), and the 
other equal angle at the circumference C is measured by 
half the arc BD ; therefore the halves of the arcs AC, 
BD, and, consequently, the arcs themselves, are also equal. 
Q. E. D. 




THEOREM XVII. 

When a tangent and chord are parallel to each other, they 
intercept equal arcs. 

Let the tangent ABC be parallel to the 
chord DF ; then are the arcs BD, BF, equal ; 
thatis,BD=BF. 

Draw the chord BD. Then, because the 
lines AB, DF, are parallel, the alternate an- 
gles D and B are equal (th. 13, B. I.). But 
the angle B, formed by a tangent and chord, is measured by 
half the arc BD (th. 8), and the other angle at the circum- 
ference D is measured by half the arc BF (th. 9) ; therefore 
the arcs BD, BF are equal. Q. E. D. 




THEOREMS. 



43 



THEOREM XVHI. 

The angle formed within a circle by the intersection of two 
chords is measured by half the sum of the two intercepted arcs. 

Let the two chords AB, CD intersect at 
the point E; then the angle AEC or DEB 
is measured by half the sum of the two arcs 
AC, DB. 

Draw the chord AF parallel to CD. 
Then, because the lines AF, CD are paral- 
lel, and AB cuts them, the angles on the 
same side A and DEB are equal (th. 15, B. I.), 
angle at the circumference A is measured by half the arc 
BF, or of the sum of FD and DB (th. 9) ; therefore the an- 
gle E is also measured by half the sum of FD and DB. 

Again, because the chords AF, CD are parallel, the arcs 
AC, FD are equal (th. 16); therefore the sum -of the two 
arcs AC, DB is equal to the sum of the two FD, DB ; and, 
consequently, the angle E, which is measured by half the 
latter sum, is also measured by half the former. Q. E. D. 




But the 



THEOREM XIX. 

The angle formed out of a circle by two secants is measured 
by half the difference of the intercepted arcs. 

Let the angle E be formed by two secants 
EAB and ECD ; this angle is measured by 
half the difference of the two arcs AC, DB, 
intercepted by the two secants. 

Draw the chord AF parallel to CD. 
Then, because the lines AF, CD are paral- 
lel, and AB cuts them, the angles on the 
same side A and BED are equal (th. 15, B. I.). But the 
angle A, at the circumference, is measured by half the arc 
BF (th. 9), or of the difference of DF and DB ; therefore 
the equal angle E is also measured by half the difference of 
DF, DB. 

Again, because the chords AF, CD are parallel, the arcs 
AC, FD are equal (th. 16) ; therefore the difference of the 
two arcs AC, DB is equal to the difference of the two DF, 





44 GEOMETRY. 

DB. Consequently, the angle E, which is measured by half 
the latter difference, is also measured by half the former. 
Q. E. D. 

THEOREM XX. 

The angle formed by two tangents is measured by half the 
difference of the two intercepted arcs. 

Let EB, ED, be two tangents to a 
circle at the points A, C ; then the an- 
gle E is measured by half the difference 
of the two arcs CFA, CGA. 

Draw the chord AF parallel to ED. 
Then, because the lines AF, ED are ® 
parallel, and EB meets them, the angles on the same side A 
and E are equal (th. 15). But the angle A, formed by the 
chord AF and tangent AB, is measured by half the arc AF 
(th. 8) ; therefore the equal angle E is also measured by 
half the same arc AF, or half the difference of the arcs CFA 
and CF, or CGA (th. 17). 

Corol. In like manner it is proved that the angle formed 
by a tangent and a secant is measured by half the difference 
of the two intercepted arcs. 

THEOREM XXI. 

When two lines, meeting a circle each in two points, cut one 
another, either within it or without it, the rectangle of the parts 
of the one is equal to the rectangle of the parts of the other, 
the parts of each being measured from the point of meeting to 
the two intersections ivith the circumference. 

Let the two lines AB, CD meet each other 
in E ; then the rectangle of AE, EB will be 
equal to the rectangle of CE, ED. Or, AE . 

EBrzrCE . ED. 

For through the point E draw the diameter 
FG- ; also, from the centre H draw the radius 
DH, and draw HI perpendicular to CD. 

Then, since DEH is a triangle, and the perpendicular HI 
bisects the chord CD (th. 1), the line CE is equal to the dif- 
ference of the segments DI, EI, the sum of them being DE. 




THEOREMS. 



45 




Also, because H is the centre of the circle, and 
the radii DH, FH, GH, are all equal, the line 
EG is equal to the sum of the sides DH, HE, 
and EF is equal to their difference. 

But the rectangle of the sum and difference 
of the two sides of a triangle is equal to the ™ 
rectangle of the sum and difference of the seg- 
ments of the base (th. 35, B. I.) ; therefore the rectangle of 
FE, EG is equal to the rectangle of CE, ED. In like man- 
ner it is proved that the same rectangle of FE, EG is equal 
to the rectangle of AE, EB. Consequently, the rectangle 
of AE, EB is also equal to the rectangle of CE, ED (ax. 
1). Q. E. D. 

Corol. 1. When one of the lines in the sec- 
ond case, as DE, by revolving about the point 
E, comes into the position of the tangent EC 
or ED, the two points C and D running into 
one, then the rectangle of CE, ED becomes 
the square of CE, because CE and DE are 
then equal. Consequently, the rectangle of 
the parts of the secant AE . EB is equal to 
the square of the tangent CE 2 . 

Corol. 2. Hence both the tangents EC, EF, drawn from 
the same point E, are equal, since the square of each is 
equal to the same rectangle or quantity AE . EB. 




THEOREM XXH. 

In equiangular triangles, the rectangles of the correspondi 
or like sides, taken alternately, are equal. 

Let ABC, DEF be two equiangular 
triangles having the angle Am the angle 
D, the angle B= the angle E, and the / 
angle C = the angle F; also the like sides G|s 
AB, DE, and AC, DF, being those op- 
posite the equal angles; then will the 
rectangle of AB, DF be equal to the 
rectangle of AC, DE. 

In BA produced take AG equal to DF, and through 
the three points B, C, G conceive a circle BCGH to be de- 
scribed, meeting CA produced at H, and join GE 




46 GEOMETRY. 

Then the angle G is equal to the angle C on the same arc 
BH, and the angle H equal to the angle B on the same arc 
CG (th. 10) ; also the opposite angles at A are equal (th. 
7, B. I.) ; therefore the triangle AGH is equiangular to the 
triangle ACB, and consequently to the triangle DFE also. 
But the two like sides AG, DF are also equal by supposi- 
tion, consequently the two triangles AGH, DFE are iden- 
tical (th. 2, B. I.), having the two sides AG, AH equal to 
the two DF, DE, each to each. 

But the rectangle GA . AB is equal to the rectangle 
HA. AC (th. 21); consequently the rectangle DF. AB is 
equal to the rectangle DE . AC. Q. E. D. 

THEOREM XXIII. 

The rectangle of the two sides of any triangle, is equal to the 
rectangle of the perpendicular on the third side and the diame- 
ter of the circumscribing circle. 

Let CD be the perpendicular, and CE the 
diameter of the circle about the triangle ABC ; 
then the rectangle CA. CB is— the rectangle 
CD . CE. 

For, join BE ; then, in the two triangles 
ACD, ECB, the angles A and E are equal, standing on the 
same arc BC (th. 10) ; also the right angle D is equal to the 
angle B, which is also a right angle, being in a semicircle 
(th. 12); therefore these two triangles have also their third 
angles equal, and are equiangular. Hence AC, CE, and 
CD, CB, being like sides, subtending the equal angles, the 
rectangle AC . CB, of the first and last of them is equal to 
the rectangle CE . CD of the other two (th. 22). 

THEOREM XXIV. 

The square of a line bisecting any angle of a triangle, to- 
gether with the rectangle of the two segments of the opposite side, 
is equal to the rectangle of the two other sides including the bi- 
sected angle. 

Let CD bisect the angle C of the triangle ABC ; then 
the square CD 2 -J- the rectangle AD.DB is=:the rectangle 
AC . CB. 





THEOREMS. 47 

For, let CD be produced to meet the circum- 
scribing circle at E, and join AE. 

Then the two triangles AEC, BCD are equi- 
angular ; for the angles at C are equal by sup- -^ 
position, and the angles B and E are equal, 
standing on the same arc AC (th. 10); consequently the 
third angles at A and D are equal (Cor. 1, th. 18, B. I.) ; 
also AC, CD, and CE, CB, are like or corresponding sides, 
being opposite to equal angles ; therefore the rectangle AC . 
CB is = the rectangle CD . CE (th. 22). But the latter rect- 
angle CD . CE is=CD 2 +the rectangle CD . DE (th. 30, B. 
I.) ; therefore the former rectangle AC . CB is also = CD 2 -f- 
CD . DE, or equal to CD 2 + AD . DB, since CD . DE is= 
AD.DB(th. 21). Q. E. D. 

THEOREM XXV. 

The rectangle of the two, diagonals of any quadrangle in- 
scribed in a circle is equal to the sum of the two rectangles of 
the opposite sides. 

Let ABCD be any quadrilateral inscribed in 
a circle, and AC, BD its two diagonals ; then 
the rectangle AC . BD is=the rectangle AB . 
DC + the rectangle AD . B C. 

For, let CE be drawn, making the angle BCE equal to 
the angle DCA. Then the two triangles ACD, BCE are 
equiangular ; for the angles A and B are equal, standing on 
the same arc CD ; and the angles DCA, BCE, are equal by 
supposition ; consequently, the third angles ADC, BEC are 
also equal ; also AC, BC, and AD, BE, are like or corre- 
sponding sides, being opposite to the equal angles ; therefore 
the rectangle AC . BE is = the rectangle AD . BC (th. 22). 

Again, the two triangles ABC, DEC are equiangular; 
for the angles BAC, BDC are equal, standing on the same 
arc BC; and the angle DCE is equal to the angle BCA, 
by adding the common angle ACE to the two equal angles 
DCA, BCE ; therefore the third angles E and ABC are 
also equal ; but AC, DC, and AB, DE, are the like sides ; 
therefore the rectangle AC . DE is=the rectangle AB . DC 
(th. 22). 

Hence, by equal additions, the sum of the rectangles AC . 




48 GEOMETRY. 

BE+ AC . DE is=zAD . BC+ AB . DC. But the former 
sum of the rectangles AC . BE-f AC . DE is = the rectangle 
AC . BD (th. 30, B. I.) ; therefore the same rectangle AC . 
BD is equal to the latter sum, the rectangle AD . BC -j- the 
rectangle AB . DC (ax. 1). Q. E. D. 

Corol. Hence, if ABD be an equilateral triangle, and C 
any point in the arc BCD of the circumscribing circle, we 
have AC=BC+DC. For AC . BD being = AD . BC+ 
AB . DC ; dividing by BD=AB=AD, there results AC= 
BC+DC 



BOOK III. 
OF RATIOS AND PROPORTIONS. 



DEFINITIONS. 



Def. 78. Ratio is the proportion or relation which one 
magnitude bears to another magnitude of the same kind 
with respect to quantity. 

Note. The measure or quantity of a ratio is conceived by 
considering what part or parts the leading quantity, called 
the antecedent, is of the other, called the consequent ; or 
what part or parts the number expressing the quantity of 
the former is of the number denoting in like manner the lat- 
ter. So the ratio of a quantity expressed by the number 2 
to a like quantity expressed by the number 6, is denoted by 
2 divided by 6, or § or -^ : the number 2 being 3 times con- 
tained in 6, or the third part of it. In like manner, the ra- 
tio of the quantity 3 to 6 is measured by f- or -J ; the ratio 
of 4 to 6 is | or •§-; that of 6 to 4 is f or -§, &c. 

79. Proportion is an equality of ratios. Thus, 

80. Three quantities are said to be proportional when 
the ratio of the first to the second is equal to the ratio of 
the second to the third. As of the three quantities A (2), 
B (4), C (8), where f =|-=£, both the same ratio. 

81. Four quantities are said to be proportional when 
the ratio of the first to the second is the same as the ratio 
of the third to the fourth. As of the four A (4), B (2), C 
(10), D (5), where ^—1^=2, both the same ratio. 

Note. To denote that four quantities, A, B, C, D, are pro- 
portional, they are usually stated or placed thus : A : B : : C 
: D, and read thus : A is to B as C is to D. But when 
three quantities are proportional, the middle one is repeat- 
ed, and they are written thus : A : B : : B : C. 



50 GEOMETRY. 

The proportionality of quantities may also be expressed 
very generally by the equality of fractions. 

Thus, if —=—, then A : B :: C : D ; also B : A ::D : C, 
' B D ' 

A:C::B:D, andC:A::D:B. 

82. Of three proportional quantities, the middle one is 
said to be a mean proportional between the other two, and 
the last a third proportional to the first and second. 

83. Of four proportional quantities, the last is said to be 
a fourth proportional to the other three, taken in order. 

84. Quantities are said to be continually proportional, or 
in continued proportion, when the ratio is the same between 
every two adjacent terms, viz., when the first is to the sec- 
ond, as the second to the third, as the third to the fourth, as 
the fourth to the fifth, and so on, all in the same common 
ratio, as in the quantities, 1, 2, 4, 8, 16, &c, where the 
common ratio is equal to 2. 

85. Of any number of quantities, A, B, C, D, the ratio of 
the first, A, to the last, D, is said to be compounded of the 
ratios of the first to the second, of the second to the third, 
and so on to the last. 

86. Inverse ratio is when the antecedent is made the 
consequent, and the consequent the antecedent. Thus, if 
1 : 2 :: 3 : 6, then inversely, 2 : 1 :: 6 : 3. 

87. Alternate proportion is when antecedent is compared 
with antecedent, and consequent with consequent. As, if 
1 : 2 : : 3 : 6, then, by alternation or permutation, it will be 
1:3:: 2:6. 

88. Compound ratio is when the sum of the antecedent 
and consequent is compared either with the consequent or 
with the antecedent. Thus, if 1 : 2 : : 3 : 6, then, by compo- 
sition, 1 + 2 : 1 :: 3 + 6 : 3, and 1 + 2 : 2 ::3 + 6 : 6. 

89. Divided ratio is when the difference of the antecedent 
and consequent is compared either with the antecedent or 
with the consequent. Thus, if 1 : 2 : : 3 : 6, then, by divis- 
ion, 2-1: 1::6— 3:3, and 2 — 1 : 2 :: 6— 3 : 6. 

Note. — The term Divided, or Division, here means sub- 
tracting or parting, being used in the sense opposed to com- 
pounding or adding, in Def. 88. 



THEOREMS. 51 



THEOREM I. 



Equimultiples of any two quantities have the same ratio as 
the quantities themselves. 

Let A and B be any two quantities, and mA, mB, any 
equimultiples of them, m being any number whatever ; then 
will raA and mB have the same ratio as A and B, or A : B 

: : mA : mB. 

„ mB B + , 

lor — =_ , the same ratio. 
mA A 

Corol. Hence like parts of quantities have the same ra- 
tio as the wholes, because the wholes are equimultiples of 
the like parts, or A and B are like parts of mA and mB, 

THEOREM II. 

If four quantities of the same kind be proportionals, they will 
be in proportion by alternation or permutation, or the antece- 
dents will have the same ratio as the consequents. 

Let A : B : : mA : mB ; then will A : mA. : : B : mB. 

_^ mA m _ mB m , , ' 
r or — — =—, and -^-=rr, both the same ratio. 
A 1 B 1 

THEOREM ni. 

If four quantities be proportional, they will be in proportion 
by inversion or inversely. 

Let A : B : : mA : mB ; then will B ; A : : mB : mA. 

mA. A. 
For — — =— , both the same ratio. 
mB B 

Otherwise. LetA:B::C:D; then shall B: A:: D: C. 

A C 

For, let — :=— —r ; then A=zBr, and C—T>r; therefore 
B D 

A C* BIT)! 

B— — , and D=— . Hence -r=-, and 7^ = -. Whence it 
r r A r C r 

is evident that -r=j^ (ax. 1), or B : A :: D : C. 

In a similar manner may most of the other theorems be 
demonstrated. 



52 GEOMETRY. 



THEOREM IV. 

If four quantities be proportional, they will be in proportion 
by composition and division. 

Let A : B : : mA : mB. 

Then will B ± A : A : : niB dzmA : mA, 
and B±A:B::raB±mA: mB. 

_ mA A , mB B 

-bor-^r r=^ r; and 



mB±mA B±A '" mBzkmA B±A* 
Corol It appears from hence that the sum of the great- 
est and least of four proportional quantities of the same kind 
exceeds the sum of the other two. For, since A: A+B 
: : mA : : mA+B — where A is the least, and mA+wB the 
greatest ; then (m+ 1) . A-j-mB, the sum of the greatest and 
least, exceeds (m-J-l).A-|-B, the sum of the two other 
quantities. 

THEOREM V. 

If of four proportional quantities there be taken any equi- 
multiples whatever of the two antecedents, and any equimul- 
tiples whatever of the two consequents, the quantities resulting 
will still be proportional. 

Let A : B : : mA : mB ; also, let pA and pmA be any equi- 
multiples of the two antecedents, and qB and qmB any equi- 
multiples of the two consequents ; then will pA : qB 

: :pmA : qniB. 

For - — -=.=—-, both the same ratio. 
pmA pA 

THEOREM VI. 

If there be four proportional quantities, and the two conse- 
quents be either augmented or diminished by quantities that have 
the same ratio as the respective antecedents, the results and the 
antecedents will still be proportionals. 

Let A : B : : mA : mB, and nA and nmA any two quanti- 
ties having the same ratio as the two antecedents; then 
will A : B±nA : : mA : mBzk.nmA. 

_ mB±nmA B±?iA , _ ■ 

For = r — , both the same ratio. 

mA A 



THEOREMS. 53 



THEOREM VII. 

If any number of quantities be proportional, then any one of 
the antecedents will be to its consequent as the sum of all the an- 
tecedents is to the sum of all the consequents. 

Let A : B : : mA : rriB : : nA : riB, &c. ; then will A : B : : A 

H-mA+^A: B+mB+nB, &c. 

^ B+mB + wB (l+m+ri)B B , 

For ' . A r =~ f-r- =— ■, the same ratio. 

A+mA+nA (l+m+«)A A' 

THEOREM VIII. 

If a whole magnitude be to a whole as apart taken from the 
first is to a part taken from the other, then the remainder will 
be to the remainder as the whole to the whole. 

LetA:B::-A:-B; 

n n 

rm rm 

Then will A : B : : A--A : B- B. 
n n 

B-=B 

n B 

For =— both the same ratio. 

A-™A A 

n 

THEOREM IX. 

If any quantities be proportional, their squares, or cubes, or 
any like powers or roots of them will also be proportional. 

Let A : B : : mA : raB ; then will A n : B n : : m 1l A n : m n B\ 

_ m n B n B n , , , 

For — — =— -, both the same ratio. 



See also theorem 8. 



THEOREM X. 

If there be two sets of proportionals, then the products or 
rectangles of the corresponding terms will also be proportional* 

Let A : B : : mA : mB, 
and C:B::nC:nT>; 
then will AC : BD : : mnAC : mnBD. 



54 GEOMETRY. 

^ rnnBD BD , , , 

For m^-rT^i both the same ratio. 

mnAC AC 

THEOREM XI. 

If four quantities be proportional, the rectangle or product 
of the two extremes will be equal to the rectangle or product of 
the two means. And the converse. 

Let A : B : : mA : mB ; 
then is A x mB=B x mA=mAB, as is evident. 

THEOREM Xn. 

If three quantities be continued proportionals, the-rectangle or 
product of the two extremes will be equal to the square of the 
mean. And the converse. 

Let A, mA, m 2 A, be three proportionals ; 

or A : mA : : mA : m 2 A ; 

then is Axm 2 A=m 2 A 2 , as is evident. 

THEOREM XIII. 

If any number of quantities be continued proportionals, the 
ratio of the first to the third will be duplicate or the square of 
the ratio of the first and second; and the ratio of the first and 
fourth will be triplicate or the cube of that of the first and sec- 
ond ; and so on. 

Let A, mA, m 2 A, m 3 A, &c., be proportionals ; 

A 1 A 1 A 1 

then is — T =— ; but ——-=—; and -—=—, &c. 

mA m m 2 A m 2 m s A m d 



BOOK IV. 




THEOREM I. 

Triangles, and also parallelograms, having equal altitudes, are 
to each other as their bases. 

Let the two triangles ADC, DEF, have the J c K 
same altitude, or be between the same paral- 
lels AE, IF ; then is the surface of the trian- 
gle ADC to the surface of the triangle DEF 
as the base AD is to the base DE ; or AD : 
DE : : the triangle ADC : the triangle DEF. 

For, let the base AD be to the base DE as any one num- 
ber m (2) to any other number n (3), and divide the re- 
spective bases into those parts, AB, BD, DG, GH, HE, all 
equal to one another, and from the points of division draw 
the lines BC, GF, HF to the vertices C and F ; then will 
these lines divide the triangles ADC, DEF, into the same 
number of parts as their bases, each equal to the triangle 
ABC, because those triangular parts have equal bases and 
altitude (Cor. 2, th. 25, B. I.) ; namely, the triangle ABC 
equal to each of the triangles BDC, DFG, GFH, HFE. 
So that the triangle ADC is to the triangle DFE as the 
number of parts m (2) of the former to the number n (3) of 
the latter ; that is, as the base AD to the base DE (def. 81). 

In like manner, the parallelogram ADKI is to the par- 
allelogram DEFK as the base AD is to the base DE, each 
of these having the same ratio as the number of their parts, 
m to n. Q. E. D. 



theorem n. 

Triangles, and also parallelograms, having equal 
each other as their altitudes. 



, are to 



Let ABC, BEF, be two triangles having the equal bases 



56 



GEOMETRY. 




AB, BE, and whose altitudes are the per- 
pendiculars CG-, FH; then will the tri- 
angle ABC : the triangle BEF : : CG : FH. 

For, let BK be perpendicular to AB, 
and equal to CG ; in which let there be 
taken BL=FH, drawing AK and AL. 

Then, triangles of equal bases and heights being equal 
(Cor. 2, th. 25, B. I.), the triangle ABK is=ABC, and 
the triangle ABL=BEF. But, considering now ABK, 
ABL, as two triangles on the bases BK, BL, and having 
the same altitude AB, these will be as their bases (th. 1), 
namely, the triangle ABK : the triangle ABL : : BK : BL. 

But the triangle ABK=ABC, and the triangle ABL= 
BEF; also BK=CG, and BL=FH. 

Therefore the triangle ABC : triangle BEF : : CG : FH. 

And since parallelograms are the doubles of these trian- 
gles, having the same bases and altitudes, they will likewise 
have to each other the same ratio as their altitudes, 
Q. E. D. 

Corol. Since, by this theorem, triangles and parallelo- 
grams, when their bases are equal, are to each other as 
their altitudes ; and, by the foregoing one, when their alti- 
tudes are equal, they are to each other as their bases; 
therefore, universally, when neither are equal, they are to 
each other in the compound ratio, or as the rectangle or 
product of their bases and altitudes. ' 



THEOREM ni. 

If four lines be proportional, the rectangle of the extremes 
will be equal to the rectangle of the means ; and, conversely, if 
the rectangle of the extremes of four lines be equal to the rect- 
angle of the means, the four lines, taken alternately, will be 
proportional. 

Let the four lines A, B, C, D be pro- 
portionals, or A : B : : C : D ; then will the 
rectangle of A and D be equal to the rect- 
angle of B and C ; or the rectangle A.D 
=B.C. 

For, let the four lines be placed with their four extrem- 
ities meeting in a common point, forming at that point four 



B 






] 

A 


K. 


DP| 



THEOREMS. 57 

right angles ; and draw lines parallel to them to complete 
the rectangles P, Q, R, where P is the rectangle of A and 
D, Q the rectangle of B and C, and R the rectangle of B 
andD. 

Then the rectangles P and R, being between the same 
parallels, are to each other as their bases A and B (th. 1) ; 
and the rectangles Q and R, being between the same paral- 
lels, are to each other as their bases C and D. But the 
ratio of A to B is the same as the ratio of C to D, by hy- 
pothesis ; therefore the ratio of P to R is the same as the 
ratio of Q to R ; and, consequently, the rectangles P and 
Q are equal. Q. E. D. 

Again, if the rectangle of A and D be equal to the rect- 
angle of B and C, these lines will be proportional, or A : B 
::C:D. 

For, the rectangles being placed the same as before, then, 
because parallelograms between the same parallels are to 
one another as their bases, the rectangle P : R : : A : B, and 
Q : R : : C : D. But, as P and Q are equal, by supposi- 
tion, they have the same ratio to R ; that is, the ratio of 
A to B is equal to the ratio of C to D, or A : B : : C : D, 
Q. E. D. 

Corol. 1. When the two means, namely, the second and 
third terms, are equal, their rectangle becomes a square of 
the second term, which supplies the place of both the sec- 
ond and third. And hence it follows that when three lines 
are proportionals, the rectangle of the two extremes is equal 
to the square of the mean ; and, conversely, if the rectan- 
gle of the extremes be equal to the square of the mean, the 
three lines are proportionals. 

Corol. 2. Since it appears by the rules of proportion in 
arithmetic and algebra that when four quantities are pro- 
portional, the product of the extremes is equal to the prod- 
uct of the two means ; and, by this theorem, the rectangle 
of the extremes is equal to the rectangle of the two means ; 
it follows that the area or space of a rectangle is represent- 
ed or expressed by the product of its length and breadth 
multiplied together. And, in general, a rectangle in geom- 
etry is similar to the product of the measures of its two di- 
mensions of length and breadth, or base and height. Also, 
a square is similar to, or represented by, the measure of its 

C2 



58 GEOMETRY. 

side multiplied by itself. So that what is shown of such 
products is to be understood of the squares and rectangles. 

Corol. 3. Since the same reasoning as in this theorem 
holds for any parallelograms whatever, as well as for the 
rectangles, the same property belongs to all kinds of par- 
allelograms, having equal angles, and also to triangles, which 
are the halves of parallelograms ; namely, that if the sides 
about the equal angles of parallelograms or triangles be re- 
ciprocally proportional, the parallelograms or triangles will 
be equal ; and, conversely, if the parallelograms or trian- 
gles be equal, their sides about the equal angles will be re- 
ciprocally proportional. 

Corol. 4. Parallelograms or triangles, having an angle in 
each equal, are in proportion to each other as the rectangles 
of the sides which are about these equal angles. 

THEOREM IV. 

If a line be draivn in a triangle parallel to one of its sides, it 
will cut the other two sides proportionally. 

Let DE be parallel to the side BC of the A 

triangle ABC; then will AD:DB::AE: 
EC. „ 




For, draw BE and CD ; then the trian- 
gles DBE, DCE, are equal to each other, 
because they have the same base DE, and B C 

are between the same parallels DE, BC (th. 25, B. I.). But 
the two triangles ADE, BDE, on the bases AD, DB, have 
the same altitude ; and the two triangles ADE, CDE, on the 
bases AE, EC, have also the same altitude; and because 
triangles of the same altitude are to each other as their bases, 
therefore 

the triangle ADE : BDE : : AD : DB, 
and triangle ADE : CDE : : AE : EC. 

But BDE is = CDE, and equals must have to equals the 
same ratio ; therefore AD : DB : : AE : EC Q. E. D. 

Corol. Hence, also, the whole lines AB, AC are propor- 
tional to their corresponding proportional segments (Cor., 
th. 1, B. III.), 

viz., AB:AC::AD:AE, 

and AB:AC::BD:CE. 




THEORE3IS. 59 



THEOREM V. 

A line which bisects any angle of a triangle divides the op- 
posite side into two segments, which are proportional to the two 
other adjacent sides. 

Let the angle CAB of the trian- JJ 
gle ABC be bisected by the line AD, \ 
making the angle r equal to the an- 
gle s ; then will the segment CD be 
to the segment DB as the side AC is 
to the side AB ; or CD : DB : : AC 
:AB. 

For, let BE be parallel to AD, meeting CA produced at 
E ; then, because the line BA cuts the two parallels AD, 
BE, it makes the angle ABE equal to the alternate angle s 
(th. 13, B. I.), and therefore also equal to the angle r, which 
is equal to s by the supposition. Again, because the line CE 
cuts the two parallels DA, BE, it makes the angle E equal 
to the angle r on the same side of it (th. 15, B. I.). Hence, 
in the triangle BAE, the angles B and E, being each equal 
to the angle r, are equal to each other, and, consequently, 
their opposite sides AB, AE are also equal (th. 3, B. I.). 

But now, in the triangle CBE, the line AD, being drawn 
parallel to the side BE, cuts the two other sides CB, CE 
proportionally (th. 4), making CD to DB as is AC to AE, 
or to its equal AB. Q. E. D. 

Case 2. The proposition is also applicable when an ex- 
ternal angle of a triangle is bisected. 

Let AC, one of the sides of the trian- 
gle ABC, be produced to E, and let the 
angle BCE be bisected by the straight line 
CD, cutting AB produced in D ; then B - 

AD:DB::AC:CB. 

Let BF be parallel to CD. 

Then, because the line BC cuts the parallel lines CD, FB, 
it makes the angle CBF equal to the alternate angle BCD ; 
and, therefore, also equal to the angle DCE, which is equal 
to BCD by supposition. Again, because the line EA cuts 
the two parallel lines CD, FB, it makes the angle DCE 
equal to the angle CFB, on the same side of the line. 



60 GEOMETRY. 

Hence, in the triangle BCF, the angles BFC and FBC, be- 
ing each equal to the angle DCE, are equal to each other ; 
and, consequently, their opposite sides BC, CF are also equal. 

Now, in the triangle ADC, the line BF, being drawn par- 
allel to the side CD, cuts the two sides AD, AC proportion- 
ally, making 

AD : AC : : DB : CF (th. 7, B. HI.). 

Or AD:DB::AC:CF. 

But BC is equal to CF ; therefore, 

AD:DB::AC:CB. 

THEOREM VI. 

Equiangular triangles are similar, or have their like sides pro- 
portional. 

Let ABC, DEF, be two equiangular tri- 
angles, having the angle A equal to the an- 
gle D, the angle B to the angle E, and, con- 
sequently, the angle C to the angle F ; then ,, 
willAB:AC::DE:DF. 

For, make DG=zAB, and DH=AC, and 



join GH; then the two triangles ABC, 
DGH, having the two sides AB, AC equal to the two DG, 
DH, and the contained angles A and D also equal, are iden- 
tical, or equal in all respects (th. 1, B. I.) ; namely, the an- 
gles B and C are equal to the angles G and H. But the 
angles B and C are equal to the angles E and F by the hy- 
pothesis ; therefore, also, the angles G and H are equal to 
the angles E and F (ax. 1), and, consequently, the line GH 
is parallel to the side EF (Cor. 1, th. 15, B. I.). 

Hence, then, in the triangle DEF, the line GH, being 
parallel to the side EF, divides the two other sides propor- 
tionally, making DG : DH : : DE : DF (Cor., th. 4). But DG 
and DH are equal to AB and AC ; therefore, also, AB : AC 
::DE:DF, Q, E, Dr 



THEOREMS. 61 



THEOREM VH. 




Triangles which have their sides proportional are also equi- 
angular. 

In the two triangles ABC, DEF, if AB : DE : : AC : DF 

: : BC : EF ; the two triangles will have their 
corresponding angles equal. 

For if the triangle ABC be not equiangu- 
lar with the triangle DEF, suppose some oth- 
er triangle, as DEG-, to be equiangular with 
ABC. But this is impossible ; for if the two 
triangles ABC, DEG were equiangular, their 
sides would be proportional (th. 6). So that, 
AB being to DE as AC to DG, and AB to D 
DE as BC to EG, it follows that DG and EG, being fourth 
proportionals to the same three quantities, as well as the 
two DF, EF, the former, DG, EG, would be equal to the 
latter, DF, EF. Thus, then, the two triangles DEF, 
DEG, having their three sides equal, would be identical 
(th. 5, B. I.), which is absurd, since their angles are une- 
qual. 

THEOREM Vm. 

Triangles which have an angle in the one equal to an angle 
in the other ; and the sides about these angles proportional, are 
equiangular. 

Let ABC, DEF, be two triangles, having the angle A= 
the angle D, and the sides AB, AC proportional to the sides 
DE, DF ; then will the triangle ABC be equiangular with 
the triangle DEF. 

For, make DG=AB, and DH=AC, and join GH. 

Then the two triangles ABC, DGH, having two sides 
equal, and the contained angles A and D equal, are iden- 
tical and equiangular (th. 1, B. I.), having the angles G and 
H equal to the angles B and C. But, since the sides DG, 
DH are proportional to the sides DE, DF, the line GH is 
parallel to EF (th. 4) ; hence the angles E and F are equal 
to the angles G and H (th. 15, B. I.), and, consequently, to 
their equals B and C. Q. E. D. [See fig. th. 6.] 




62 GEOMETRY. 



THEOREM IX. 

In a right-angled triangle, a perpendicular from the right 
angle is a mean proportional between the segments of the hy- 
pothenuse, and each of the sides about the right angle is a 
mean proportional between the hypothenuse and the adjacent 
segment. 

Let ABC be a right-angled triangle, and 
CD a perpendicular from the right angle C 
to the hypothenuse AB ; then will £ Jo~~B 

CD be a mean proportional between AD and DB ; 
AC a mean proportional between AB and AD ; 
BC a mean proportional between AB and BD. 

Or AD:CD::CD:DB; and AB:BC ::BC : BD; and 
AB:AC::AC:AD. 

For the two triangles ABC, ADC, having the right an- 
gles at C and D equal, and the angle A common, have their 
third angles equal, and are equiangular (Cor. 1, th. 18, B. I.). 
In like manner, the two triangles ABC, BDC, having the 
right angles at C and D equal, and the angle B common, 
have their third angles equal, and are equiangular. 

Hence, then, all the three triangles, ABC, ADC, BDC, 
being equiangular, will have their like sides proportional 
(th.6), 

viz., AD:CD::CD:DB; 

and AB:AC::AC:AD; 

and AB:BC::BC:BD. Q. E. D. 

Corol. 1. Because the angle in a semicircle is a right an- 
gle (th. 12, B. II.), it follows that, if from any point C in 
the periphery of the semicircle a perpendicular be drawn to 
the diameter AB, and the two chords CA, CB be drawn to 
the extremities of the diameter, then are AC, BC, CD the 
mean proportionals as in this theorem, or (by th. 12, B. III.), 
CD 2 =AD.DB; AC 2 = AB.AD; and BC 2 =AB.BD. 

Corol 2. Hence AC 2 : BC 2 : : AD : BD. 

Corol. 3. Hence we have another demonstration of th. 
34, B. I. 

For, since AC 2 =AB. AD, and BC 2 =AB.BD; 

By addition, AC 2 +BC 2 =AB(AD-f BD)=AB 2 . 



THEOREMS. 63 



THEOREM X. 




Equiangular or similar triangles are to each other as the 
squares of their like sides. 

Let ABC, DEF be two equiangular tri- 
angles, AB and DE being two like sides; 
then will the triangle ABC be to the trian- 
gle DEF as the square of AB is to the square 
of DE, or as AB 2 to DE 2 . 

For, the triangles being similar, they have 
their like sides proportional (th. 6), and are to G ^ 

each other as the rectangles of the like pairs of their sides 
(Cor. 4, th. 3) ; 

therefore AB : DE : : AC : DF (th. 6), 
and AB : DE : : AB : DE of equality ; 

therefore AB 2 : DE 2 : : AB . AC : DE . DF (th. 10, B. IH). 

But A ABC : A DEF : : AB . AC : DE . DF (Cor. 4, th. 3) ; 
therefore A ABC : A DEF : : AB 2 : DE 2 . Q. E. D. 

THEOREM XI. 

All similar figures are to each other as the squares of their 
like sides. 

Let ABCDE, FGHIK be 

any two similar figures, the 
like sides being AB, FG-, and 
BC, GH, and so on in the 
same order ; then will the fig- 
ure ABCDE be to the figure " EG 
FGHIK as the square of AB to the square of FG, or as 
AB 2 to FG 2 . 

For, draw BE, BD, GK, GI, dividing the figures into an 
equal number of triangles by lines from two equal angles B 
and G. 

The two figures being similar (by supposition), they are 
equiangular, and have their like sides proportional (def. 67.) 

Then, since the angle A is — the angle F, and the sides 
AB, AE proportional to the sides FG, FK, the triangles 
ABE, FGK are equiangular (th. 8). In like manner, the 
two triangles BCD, GHI, having the angle C = the angle 




64 GEOMETRY,. 

H, and the sides BC, CD proportional to the sides GH, HI, 
are also equiangular. Also, if from the equal angles AED, 
FKI, there be taken the equal angles AEB, FKG, there 
will remain the equals BED, GKI ; and if from the equal 
angles CDE, HIK be. taken away the equals CDB, HIG, 
there will remain the equals BDE, GIK ; so that the two 
triangles BDE, GIK, having two angles equal, are also equi- 
angular. Hence each triangle of the one figure is equian- 
gular with each corresponding triangle of the other. 

But equiangular triangles are similar, and are proportion- 
al to the squares of their like sides (th. 10). 

Therefore the A ABE : A FGK : : AB 2 : FG 2 , 
and A BCD : A GHI : : BC 2 : GH 2 , 

and A BDE : A GIK : : DE 2 : IK 2 . 

But as the two polygons are similar, their like sides are 
proportional, and, consequently, their squares also propor- 
tional ; so that all the ratios AB 2 to FG 2 , and BC 2 to GH 2 , 
and DE 2 to IK 2 , are equal among themselves, and, conse- 
quently, the corresponding triangles also, ABE to FGK, 
and BCD to GHI, and BDE to GIK, have all the same ra- 
tio, viz., that of AB 2 to FG 2 ; and hence all the antecedents, 
or the figure ABCDE, have to all the consequents, or the 
figure FGHIK, still the same ratio, viz., that of AB 2 to 
FG 2 (th. 7, B. III.). Q. E. D. 

THEOREM XII. 

Similar figures inscribed in circles have their like sides, and 
also their whole perimeters, in the same ratio as the diameters 
of the circles in ivhich they are inscribed. 

Let ABCDE, FGHIK be 
two similar figures inscribed 
in the circles whose diam- 
eters are AL and FM; then 
will each side AB, BC, &c, 
of the one figure be to the 
like side FG, GH, &c, of the 
other figure, or the whole perimeter AB-j-BC-f-, &c. of the 
one figure, to the whole perimeter FG+GH-f-, &c. of the 
other figure, as the diameter AL to the diameter FM. 

For, draw the two corresponding diagonals, AC. FH, as 




THEOREMS. 65 

also the lines BL, GM. Then, since the polygons are sim- 
ilar, they are equiangular, and their like sides have the same 
ratio (def. 67) ; therefore the two triangles ABC, FGH have 
the angle B=the angle G; and the sides AB, BC propor- 
tional to the two sides FG, GH ; consequently, these two 
triangles are equiangular (th. 8), and have the angle ACB 

r=FHG. But the angle ACB = ALB, standing on the same 
arc AB ; and the angle FHG-=FMG, standing on the same 
arc FG; therefore the angle ALB=rFMG (ax. 1). And, 
since the angle ABL=FGM, being both right angles, be- 
cause in a semicircle, therefore the two triangles ABL, FGM, 
having two angles equal, are equiangular ; and, consequent- 
ly, their like sides are proportional (th. 6) ; hence AB : FG 

: : the diameter AL : the diameter FM. 

In like manner, each side BC, CD, &c, has to each side 
GH, HI, &c, the same ratio of AL to FM ; and, conse- 
quently, the sums of them are still in the same ratio, viz., 
AB-f-BC + CD, &c. : FG+GH+HI, &c. :: the diam. AL: 
the diam. FM (th. 7, B. III.). Q. E. D. 

THEOREM XIII. 

Similar figures inscribed in circles are to each other as the 
squares of the diameters of those circles. 

Let ABCDE, FGHIK be two similar figures inscribed in 
the circles whose diameters are AL and FM ; then the sur- 
face of the polygon ABCDE will be to the surface of the 
polygon FGHIK as AL 2 to FM 2 . 

For the figures, being similar, are to each other as the 
squares of their like sides, AB 2 to FG 2 (th. 10). But, by 
the last theorem, the sides AB, FG are as the diameters AL, 
FM, and therefore the squares of the sides AB 2 to FG 2 as 
the squares of the diameters AL 2 to FM 2 (th. 9, B. III.). 
Consequently, the polygons ABCDE, FGHIK are also to 
each other as the squares of the diameters AL 2 to FM 2 
(ax. 1). Q. E. D. [See fig. th. 12.] 



66 GEOMETRY. 



THEOREM XIV. 



The circumferences of all circles are to each other as their di- 
ameters. 

Let D, d denote the diameters of two circles, and C, c 
their circumferences ; then will D : d : : C : c, or D : C : : d : c. 

For (by th. 12) similar polygons inscribed in circles have 
their perimeters in the same ratio as the diameters of those 
circles. 

Now, as this property belongs to all polygons, whatever 
the number of the sides may be, conceive the number of the 
sides to be indefinitely great, and the length of each infinite- 
ly small, till they coincide with the circumference of the 
circle, and be equal to it, indefinitely near ; then the perim- 
eter of the polygon of an indefinite number of sides is the 
same thing as the circumference of the circle. Hence it 
appears that the circumferences of the circles, being the same 
as the perimeters of such polygons, are to each other in the 
same ratio as the diameters of the circles. Q. E. D. 

THEOREM XV. 

The areas or spaces of circles are to each other as the squares 
of their diameters or of their radii. 

Let A, a denote the areas or spaces of two circles, and 
D, d their diameters ; then A : a : : D 2 : d?. 

For (by th. 13) similar polygons inscribed in circles are 
to each other as the squares of the diameters of the circles. 

Hence, conceiving the number of the sides of the polygons 
to be increased more and more, or the length of the sides to 
become less and less, the polygon approaches nearer and 
nearer to the circle, till at length, by an infinite approach, 
they coincide, and become in effect equal ; and then it fol- 
lows that the spaces of the circles, which are the same as 
of the polygons, will be to each other as the squares of the 
diameters of the circles. Q. E. D. 

Corol. The spaces of circles are also to each other as the 
squares of the circumferences, since the circumferences are 
in the same ratio as the diameters (by th. 14). * 



THEOREMS. 67 



THEOREM XVI. 




The area of any circle is equal to the rectangle of half its 
circumference and half its diameter. 

Conceive a regular polygon to be inscribed 
in a circle, and radii drawn to all the angular 
points, dividing it into as many equal trian- 
gles as the polygon has sides, one of which is 
ABO, of which the altitude is the perpendic- 
ular OD from the centre to the base AB. 

Then the triangle ABO, being equal to a rectangle of 
half the base and equal altitude (Cor. 2, th. 26, B. L), is 
equal to the rectangle of the half base AD and the altitude 
OD ; consequently, the whole polygon, or all the triangles 
added together which compose it, is equal to the rectangle 
of the common altitude OD, and the halves of all the sides, 
or the half perimeter of the polygon. 

Now conceive the number of sides of the polygon to be 
indefinitely increased ; then will its perimeter coincide with 
the circumference of the circle, and, consequently, the alti- 
tude OD will become equal to the radius, and the whole 
polygon equal to the circle. Consequently, the space of the 
circle, or of the polygon in that state, is equal to the rect- 
angle of the radius and half the circumference. Q. E. D. 

THEOREM XVII. 

The area of a circle is equal to the square of the radius mul- 
tiplied by the ratio of the diameter to the circumference. 

Let the circumference of the circle whose diameter is uni- 
ty be denoted by tt; then, since the diameters of circles are 
to each other as their circumferences (Prop. 14), rr will de- 
note the ratio of any circumference to its diameters. We 
shall then have (r being the radius), 

1 : it : : 2r : circumference ; 
.•. circumference =7T x 2r. 
Multiplying both members by ir, we have 
\r x circumference = 7r . r 2 ; 
or (th. 16) area=7r. r 2 . 

The numerical value of tt is 3.14159. 



(38 



GEOMETRY. 




THEOREM XVIII. 

If a straight line cut two sides of a triangle proportionally, 
this line will be parallel to the third side. (Converse of th. 4.) 

Let the line DE cut the sides AB, AC, 
so that AD : DB : : AE : EC ; then will the 
line DE be parallel to the side BC. 

Because the two triangles ADE, EDB, 
have their bases in the same straight line 
AB, and their vertices at E, they have the 
same altitude, and are to each other as their B £ 

bases (th. 1). 

Or ADE : EDB : : AD : DB. 

And, for a like reason, 

ADE:EDC::AE;EC. 

But AD:DB::AE:EC; 

.-. ADE : EDB : : ADE : EDC. 

Hence, EDB = EDC ; and, having the same base DE, 
they must have the same altitude ; 

;. DE is parallel to BC. Q. E. D. 




PROBLEMS. 



PROBLEM I. 

To make an equilateral triangle on a given line AB, 

From the centres A and B, with the distance 

AB, describe arcs intersecting in C. Draw 

AC, BC, and ABC will be the equilateral tri- 
angle. 

For the equal radii AC, BC are each of 
them equal to AB. 

PROBLEM II. 

To bisect a given angle BAC. 

From the centre A, with any radius, 
describe an arc cutting off the equal lines 

AD, AE ; and from the two centres, D, 
E, with the same radius, describe arcs in- B 
tersecting in F ; then draw AF, which 
will bisect the angle A as required. 

Join DF, EF. Then the two triangles ADF, AEF, hav- 
ing the two sides AD, DF equal to the two AE, EF (being 
equal radii), and the side AF common, they are mutually 
equilateral ; consequently, they are also mutually equian- 
gular (th. 5), and have the angle BAF equal to the angle 
CAF. 

Scholium. In the same manner is an arc of a circle bi- 
sected. 

PROBLEM in. 

To bisect a given line AB. 

From the two centres A and B, with any equal radii, 
describe arcs of circles intersecting each other in C and D, 




;s 




70 GEOMETRY. 

and draw the line CD, which will bisect the 
given line AB in the point E. 

Draw the radii AC, BC, AD, BD. Then, 
because all these four radii are equal, and the 
side CD common, the two triangles ACD, 
BCD are mutually equilateral ; consequently, 
they are also mutually equiangular (th. 5, B. 
I.), and have the angle ACE equal to the an- 
gle BCE. 

Hence the two triangles ACE, BCE, having the two sides 
AC, CE equal to the two sides BC, CE, and their contain- 
ed angles equal, are identical (th. 1), and therefore have the 
side AE equal to EB. 

PROBLEM IV. 

At a given point C, in a line AB, to erect a perpendicular. 
From the given point C, with any ra- f 



:>c 




dius, cut off any equal parts CD, CE of 

the given line ; and from the two centres 

D and E, with any one radius, describe 

arcs intersecting in F ; then join CF, A D C E~H 

which will be perpendicular as required. 

Draw the two equal radii DF, EF. Then the two tri- 
angles CDF, CEF, having the two sides CD, DF, equal to 
the two CE, EF, and CF common, are mutually equilater- 
al ; consequently, they are also mutually equiangular (th. 
5, B. I.), and have the two adjacent angles at C equal to 
each other ; therefore the line CF is perpendicular to AB 
(def. 11). 

OTHERWISE. 

When the point C is near the end of the 

From any point D, assumed above the 
line, as a centre, through the given point C 
describe a circle, cutting the given line at 
E ; and through E and the centre D, draw 
the diameter EDF ; then join CF, which 
will be the perpendicular required. 

For the angle at C, being an angle in a semicircle, is a 




/ 


>- 


\ 


/ 



PROBLEMS, 71 

right angle, and therefore the line CF is a perpendicular 
(by def. 15). 

PROBLEM V. 

From a given point A, to let fall a perpendicular on a given 
line BC. 

From the given point A as a centre, with A 

any convenient radius, describe an arc, cut- 
ting the given line at the two points D and 
E; and from the two centres D, E, with 
any radius, describe two arcs, intersecting • B ' 
at F ; then draw AGF, which will be per- 
pendicular to BC, as required. 

Draw the equal radii AD, AE, and DF, 
EF. Then the two triangles ADF, AEF, having the two 
sides AD, DF, equal to the two AE, EF, and AF common, 
are mutually equilateral ; consequently, they are also mu- 
tually equiangular (th. 5, B. I.), and have the angle DAG 
equal the angle EAG. Hence, then, the two triangles ADG, 
AEG, having the two sides AD, AG, equal to the two AE, 
AG, and their included angles equal, are therefore equian- 
gular (th. 1, B. I.), and have the angles at G equal ; conse- 
quently, AG is perpendicular to BC (def. 11). 

# OTHERWISE. 

When the point is nearly opposite the end of the line. 

From any point D, in the given line 
BC, as a centre, describe the arc of a cir- 
cle through the given point A, cutting BC 
in E ; and from the centre E, with the ra- 
dius EA, describe another arc, cutting the 
former in F ; then draw AGF, which will 
be perpendicular to BC as required. 

Draw the equal radii DA, DF, and EA, EF. Then the 
two triangles DAE, DFE, will be mutually equilateral ; 
consequently, they are also mutually equiangular (th. 5, B. 
I.), and have the angles at D equal. Hence the two tri- 
angles DAG, DFG, having the two sides DA, DG equal 
to the two DF, DG, and the included angles at D equal, 




72 GEOMETRY. 

have also the angles at G equal (tli. 1, B. I.) ; consequent- 
ly, those angles at G are right angles, and the line AG is 
perpendicular to DG-. 

PROBLEM vi. 
To make a triangle ivith three given lines, AB, AC, BC. 

With the centre A, and distance AC, describe an arc. 
With the centre B, and distance BC, describe another arc, 
cutting the former in C. Draw AC, BC, and ABC will be 
the triangle required. 

For the radii, or sides of the triangle, AC, BC, are equal 
to the given lines AC, BC, by construction. 

Note. — If any two of the lines are not together greater 
than the third, the construction is impossible. 

PROBLEM VII. 

At a given 'point A, in a line AB, to make an angle equal to 
a given angle C 

From the centres A and C, with any one 
radius, describe the arcs DE, FG. Then, 
with radius DE and centre F, describe an 
arc, cutting FG in G. Through G draw 
the line AG, and it will form the angle re- 
quired. 

Conceive the equal lines or radii, DE, FG, to be drawn. 
Then the two triangles CDE, AFG, being mutually equi- 
lateral, are mutually equiangular (th. 5, B. I.), and have the 
angle at A equal to the angle at C. 

problem vin. 

Through a given point A, to draw a line parallel .o a given 
line BC. 

From the given point A draw a line jj DC 

AD to any point in the given line BC. 
Then draw the line EAF, making the / 

angle at A equal to the angle at D -p -^ ■= 

(by Prob. 5) ; so shall EF be parallel ' A 
to BC, as required. 




PROBLEMS. 73 

For the angle D being equal to the alternate angle A, the 
lines BC, EF are parallel, by th. 13, B. I. 

PROBLEM IX. 

To divide a line AB into any proposed number of equal parts. 

Draw any other line AC, forming any 
angle with the given line AB, on which |J£ 

set off as many equal parts AD, DE, EF, ^ ^^C 



FC, as the line AB is to be divided into. - I B G 

Join BC ; parallel to which draw the other lines FG-, EH, 
DI : then these will divide AB in the manner required. 
For those parallel lines divide both the sides AB, AC pro- 
portionally, by th. 4, B. IV. 

problem x. 
To make a square on a given line AB. 

Raise AD, BC, each perpendicular and equal D c 

to AB, and join CD; so shall ABCD be the 
square sought. 

For all the three sides AB, AD, BC are equal 
by the construction, and DC is equal and parallel 
to AB (by th. 24, B. I.) ; so that all the four A B 

sides are equal, and the opposite ones are parallel. Again, 
the angle A or B of the parallelogram being a right angle, 
the angles are all right ones (Cor. 1, th. 22, B. I.). Hence, 
then, the figure, having all its sides equal and all its angles 
right, is a square (def. 34). 

PROBLEM XI. 

To make a rectangle or a parallelogram of a given length 
<ind oreadth, AB, BC. 

Erect AE, BD perpendicular to AB, and E ^ 

each equal to BC; then join DE, and it is 
done. 



The demonstration is the same as the last A. $ 
problem. ^— - 

And in the same manner is described any ** 
oblique parallelogram, only drawing AE and BD to make 

D 




74 GEOMETRY. 

the given oblique angle with AB instead of perpendicular 
to it. 

PROBLEM XII. 

To make a rectangle equal to a given triangle ABC. 

Bisect the base AB in D ; then raise DE c e f 
and BF perpendicular to AB, and meeting CF 
parallel to AB at E and F ; so shall DF be 
the rectangle equal to the given triangle ABC 
(by Cor. 2, th. 26, B. I.). 

PROBLEM Xm. 

To make a square equal to the sum of two or more given squares. 

Let AB and AC be the sides of two giv- 
en squares. Draw two indefinite lines AP, 
AQ, at right angles to each other, in which 
place the sides AB, AC of the given squares ; 
join BC ; then a square described on BC 
will be equal to the sum of the two squares 
described on AB and AC (th. 34, B. I.). 

In the same manner, a square may be made equal to the 
sum of three or more given squares. For, if AB, AC, AD 
be taken as the sides of the given squares, then, making AE 
=BC, AD=AD, and drawing DE, it is evident that the 
square on DE will be equal to the sum of the squares on 
AB, AC, AD. And so on for more squares. 

PROBLEM XIV. 

To make a square equal to the difference of two given squares. 
Let AB and AC, taken in the same straight ^ — ^r> 




line, be equal to the sides of two given squares. / /i\ 

From the centre A, with the distance AB, de- L / I \ 

scribe a circle ; and make CD perpendicular a c b 

to AB, meeting the circumference in D ; so shall a square 
described on CD be equal to AD 2 -AC 2 , or AB 2 -AC 2 , as 
required (Cor., th. 34, B. I.). 



PROBLEMS. 75 

PROBLEM XV. 

To make a triangle equal to a given quadrilateral ABCD. 

Draw the diagonal AC, and parallel d^ c 

to it DE, meeting BA produced at E, 




and join CE ; then will the triangle CEB 
be equal to the given quadrilateral 
ABCD. E A B 

For the two triangles ACE, ACD, being on the same 
base AC, and between the same parallels AC, DE, are equal 
(th. 25, B. I.) ; therefore, if ABC be added to each, it will 
make BCE equal to ABCD (ax. 2). 

PROBLEM XVI. 

To make a triangle equal to a given pentagon ABCDE. 

Draw DA and DB, and also EF, CG, 
parallel to them, meeting AB produced at 
F and G ; then draw DF and DG ; so 
shall the triangle DFG be equalto the 
given pentagon ABCDE. 

For the triangle DFA=DEA, and the 
triangle DGB=DCB (th. 25) ; therefore, by adding DAB 
to the equals, the sums are equal (ax. 2) ; that is, DAB 
+ DAF + DBG = DAB+DAE+DBC, or the triangle 
DFG — to the pentagon ABCDE. 

PROBLEM XVII. 

To make a square equal to a given rectangle ABCD. 

Produce one side AB, till BE be equal 
to the other side BC. On AE as a di- 
ameter describe a circle meeting BC pro- 
duced at F ; then will BF be the side of 
the square BFGH, equal to the given 
rectangle BD, as required, as appears by Cor., th. 9, B. IV"., 
and th. 12, B. HI. 




c 


V- — <p, 


/ 


N 


t 


p 



AH BE 




76 GEOMETRY. 



PROBLEM XVIII. 

To describe a circle about a given triangle ABC. 

Bisect any two sides with two of the per- 
pendiculars, DE, DF, DGr, and D will be 
the centre. 

For, join DA, DB, DC. Then the two 
right-angled triangles DAE, DBE, have the 
two sides DE, EA equal to the two DE, 
EB, and the included angles at E equal ; 
these two triangles are therefore identical (th. 1, B. I.), and 
have the side DA equal to DB. In like manner it is shown 
that DC is also equal to DA or DB. So that all the three, 
DA, DB, DC, being equal, they are radii of a circle passing 
through A, B, and C. 

Note. — The problem is the same in effect when it is re- 
quired 

To describe the circumference of a circle through three given 
points A, B, C. 

Then, from the point B draw chords 
BA, BC, to the two other points, and bi- 
sect these chords perpendicularly by lines 
meeting in O, which will be the centre. 
Again, from the centre O, at the distance 
of any one of the points, as OA, describe 
a circle, and it will pass through the two 
other points, B, C, as required. The demonstration is evi- 
dently as above. 

PROBLEM XIX. 

An isosceles triangle ABC being given, to describe another on 
the same base AB, whose vertical angle shall be only half the 
vertical angle C 

From C as a centre, with the distance CA, describe the 
circle ABE. Bisect AB in D, join DC, and produce to the 
circumference E ; join EA and EB, and ABE shall be the 
isosceles triangle required. 

For, since in the triangles EDA, EDB, AD is equal to 




PROBLEMS. 



77 



DB, and DE common to both, and the 
right angle EDA equal to the right angle 
EDB, the side EA must be equal to the 
side EB ; the triangle AEB is therefore 
isosceles, and the angle ACB at the centre 
must be double of the angle AEB at the 
circumference, for they both stand on the 
same segment AB. 




PROBLEM xx. 

Given an isosceles triangle AEB, to erect another on the 
hose AB, which shall have double the vertical angle E. 

Describe a circle about the triangle 
AEB ; find its centre C, and join CA, AB, 
and ACB is the triangle required. 

The angle C at the centre is double of 
the angle E at the circumference, and the 
triangle ACB is isosceles, for the sides 
CA, CB, being radii of the same circle, are 
equal. 




PROBLEM XXI. 

To find the centre of a given circle. 

Draw any chord AB, and bisect it per- 
pendicularly with the line ED ; this (Cor., 
th. 1, B. II.) will be a diameter. There- 
fore bisect ED in C, which will be the cen- 
tre, as required. 



PROBLEM XXH. 

To draw a tangent to a circle through a given point A. 

1. When the given point A is in the cir- b a. 
cumference of the circle, join A and the cen- 
tre O, perpendicular to which draw BAC, 
and it will be the tangent, by th. 6, B. LT. 

2. When the given point A is out of the 
circle, draw AO to the centre, on which, as a diameter, de- 





78 



GEOMETRY. 



scribe a semicircle, cutting the given circum- BA^ 
ference in D, through which draw BADC, 
which will be the tangent as required. 

For, join DO. Then the angle ADO, in a 
semicircle, is a right angle, and, consequently, 
AD is perpendicular to the radius DO, or is a tangent to 
the circle (th. 6, B. II.). 




PROBLEM XXIII. 

On a given line AB to describe a segment of a circle, to con- 
tain a given angle C 

At the ends of the given line make angles 
DAB, DBA, each equal to the given angle 
C. Then draw AE, BE, perpendicular to 
AD, BD ; and with the centre E, and radius 
EA, or EB, describe a circle ; so shall AFB 
be the segment required, as any angle F made 
in it will be equal to the given angle C. 

For, the two lines AD, BD, being per- 
pendicular to the radii EA, EB (by construction), are tan- 
gents to the circle (th. 6, B. II.) ; and the angle A or B, 
which is equal to the given angle C by construction, is equal 
to the angle F in the alternate segment AEB (th. 13, B. II.). 




PROBLEM XXIV. 

To cut off a segment from a circle that shall contain a given 
angle C 

Draw any tangent AB to the given cir- 
cle, and a chord AD to make the angle 
DAB equal to the given angle C ; then 
DEA will be the segment required, any an- 
gle E made in it being equal to the given 
angle C. 

PROBLEM XXV. 

To inscribe an equilateral triangle in a given circle. 

Through the centre C draw any diameter AB. From 
the point B as a centre, with the radius BC of the given 




PROBLEMS. 



79 




circle, describe an arc DCE. Join AD, 
AE, DE, and ADE is the equilateral 
sought. 

Join DB, DC, EB, EC Then DCB is 
an equilateral triangle, having each side 
equal to the radius of the given circle. In 
like manner, BCE is an equilateral trian- 
gle. But the angle ADE is equal to the angle ABE or 
CBE, standing on the same arc AE ; also the angle AED 
is equal to the angle CBD, on the same are AD ; hence the 
triangle DAE has two of its angles, ADE, AED, equal to 
the angles of an equilateral triangle, and therefore the third 
angle at A is also equal to the same, so that the triangle 
is equiangular, and therefore equilateral. 



PROBLEM XXVI. 

To inscribe a circle in a given triangle ABD. 

Bisect any two angles C and B 
with the two lines CD, BD. From 
the intersection D, which will be the 
centre of the circle, draw the per- 
pendiculars DE, DF, DGr, and they 
will be the radii of the circle re- 
quired. 

For, since the angle DCF is equal 
to the angle DCG-, and the angles B 3T C 

at F, G, right angles (by construction), the two triangles 
CDF, CDG- are equiangular ; and, having also the side CD 
common, they are identical, and have the sides DE, DGr 
equal (th. 2, B. I.). In like manner it is shown that DF is 
equal to DE or DG-. 

Therefore, if with the centre D, and distance DE, a circle 
be described, it will pass through all the three points E, F, 
G, in which points also it will touch the three sides of the 
triangle (th. 6, B. II.), because the radii DE, DF, DG, are 
perpendicular to them. 




80 



GEOMETKY. 



PROBLEM XXVII. 

To inscribe a square in a given circle. 

Draw two diameters AC, BD, crossing 
at right angles in the centre E. Then 
join the four extremities A, B, C, D, with 
right lines, and these will form the in- 
scribed square ABCD. 

For the four right-angled triangles AEB, 
BEC, CED, DEA, are identical, because 
they have the sides EA, EB, EC, ED all equal, being radii 
of the circle, and the four included angles at E all equal, be- 
ing right angles, by the construction. Therefore all their 
third sides AB, BC, CD, DA, are equal to one another, and 
the figure ABCD is equilateral. Also, all its four angles, 
A, B, C, D, are right ones, being angles in a semicircle. 
Consequently the figure is a square. 




U2 



m 



PROBLEM XXVm. 

To describe a square about a given circle. 

Draw two diameters AC, BD, crossing at j^ 
right angles in the centre E. Then through 
their four extremities draw FG, IH, parallel 
to AC, and FI, G-H, parallel to BD, and 
they will form the square FGHI. 

For, the opposite sides of parallelograms 
being equal, FG and IH are each equal to the diameter AC, 
FI and GH each equal to the diameter BD, so that the 
figure is equilateral. Again, because the opposite angles of 
parallelograms are equal, all the four angles F, G, H, I are 
right angles, being equal to the opposite angles at E. So 
that the figure FGHI, having its sides equal, and its angles 
right ones, is a square, and its sides touch the circle at the 
four points A, B, C, D, being perpendicular to the radii 
drawn to those points. 



PROBLEMS. 81 

PROBLEM XXIX. 

To inscribe a circle in a given square. 

Bisect the two sides FI, FG, in the points A and B (last 
fig.). Then, through these two points draw AC parallel to 
FG or IH, and BD parallel to FI or GIL Then the point 
of intersection E will be the centre, and the four lines EA, 
EB, EC, ED, radii of the inscribed circle. 

For, because the four parallelograms EF, EG, EH, EI, 
have their opposite sides and angles equal, therefore all the 
four lines EA, EB, EC, ED are equal, being each equal to 
half a side of the square. So that a circle described from 
the centre E, with the distance EA, will pass through all 
the points A, B, C, D, and will be inscribed in the square, 
or will touch its four sides in those points, because the an- 
gles there are right ones. 

PROBLEM XXX. 

To describe a circle about a given square. 

(See figure, Problem xxvii.) 

Draw the diagonals AC, BD, and their intersection E 
will be the centre. 

As the diagonals of a square bisect each other (th. 40), 
then will EA, EB, EC, ED be all equal, and, consequently, 
these are radii of a circle passing through the four points 
A, B, C, D. 

PROBLEM XXXI. 

To find a third proportional to two given lines, AB, AC 

Place the two given lines AB, AC (or A . 
two lines equal td them) to form any angle a.- 
at A; and in AB set off AD = AC. Join 
BC, and draw DE parallel to it ; so will 
AE, on the line AC, be the third propor- 
tional sought. 

For, since DE is parallel to BC, the two lines AB, AC 
are cut proportionally by DE (th. 4, B. IV.) ; hence AB : 
AC :: AD (=AC) : AE, and AE is, therefore, the third pro- 
portional required. 

D2 





82 GEOMETRY. 

PROBLEM XXXII. 

To find a fourth proportional to three given lines, AB, AC, AD. 

Place two of the given lines, AB, AC, ^ - £ 

or their equals, to make any angle at A ; 
and on AB set off, or place the other line 
AD, or its equal. Join BC, and parallel 
to it draw DE ; so shall AE be the fourth 
proportional as required. n 

For, because of the parallels BC, DE, the two sides AB, 
AC, are cut proportionally (th. 4, B. IV.) ; so that AB : AC 
::AD:AE. 

PROBLEM XXXIII. 

To find a mean proportional between two lines, AB, BC 

Place AB, BC, joined in one straight line A B 

AC ; on which, as a diameter, describe the B — c 

semicircle ADC ; to meet which erect the , ,*'" 
perpendicular BD, and it will be the mean / \ 

proportional sought between AB and BC, L L_J 

by Cor., th. 9, B. IV. A ° B , c 

PROBLEM XXXIV. 

To divide a given line in extreme and mean ratio. 

Let AB be the given line to be divided in ,& 

extreme and mean ratio, that is, so that the D / 

whole line may be to the greater part as the S — y^~ E 
greater is to the less part. f a [__\ tt 

Draw BC perpendicular to AB, and equal I / 7 
to half AB. Join AC ; and with centre C ^<L-*S 
and distance CB, describe the circle BD ; then, ^ 
with centre A and distance AD, describe the arc DE ; so 
shall AB be divided in E in extreme and mean ratio, or so 
that AB:AE::AE:EB. 

Produce AC to the circumference at F. Then, ADF be- 
ing a secant, and AB a tangent, because B is a right angle ; 
therefore the rectangle AF . AD is equal to AB 2 (Cor. 1, th. 
21) ; consequently the means and extremes of these are pro- 



PROBLEMS. 83 

portional (th. 12, B. HI.), viz., AB : AF or AD-j-DF : : AD : 
AB. But AE is equal to AD by construction, and AB = 
2BC=DF ; therefore AB : AE+ AB : : AE : AB or AE+ 
EB ; and by division, AB : AE : : AE : EB. 

PROBLEM XXXV. 

To cut a given line AB in a point F, so that the square of 
the one part BF may be equal to the rectangle of the whole line 
AB and the other part AF. 

Produce AB till BC be equal to it ; erect A 

the perpendicular BD equal to AB or BC ; p [C ^~ 

bisect BC in E; join ED, and make EF Ii — gV — 
equal to it ; the square of the segment BF - — ■- — - — -^ — 
is equivalent to the rectangle contained by 
the whole BA and its remaining segment AF. The line 
AB is then said to be divided by medial section at the 
point F. 

For on BC construct the square BG ; make BH equal to 
BF, and draw IHK and FI parallel to AC and BD. Since 
AB is equal to BD, and BF to BH, the remainder AF is 
equal to HD ; and it is further evident that FH is a square, 
and IC and DK are rectangles. But BC being bisected in 
E and produced to F, the rectangle under BF, FC, or the 
rectangle IC, together with the square of BE, is equivalent 
to the square of EF or DE. But the square of DE is equiv- 
alent to the squares of DB and BE ; whence the rectangle 
IC, with the square of BE, is equivalent to the squares of 
DB and BE ; or, omitting the common square of BE, the 
rectangle IC is=:to the square of DB. Take away from 
both the rectangle BK, and there remains the square BI or 
the square of BFzz:to the rectangle HG-, or the rectangle 
contained by BA and AF. 

Cor. Hence also the construction of another problem of 
the same nature, in which it is required to produce a 
straight line AB, such that the rectangle contained by the 
whole line thus produced and the part produced shall be 
equivalent to the square of the line AB itself. 

Bisect AB in C ; draw the perpendicular BD=BC ; join 
AD, and continue it until DE=DB or BC, and on AB pro- 
duced take AF=AE; the line AF is the required exten- 




C B J} 



84 GEOMETRY. 

sion of AB. For, make DG=DB or BC; D> * 

and because the rectangle EA, AG, together 

with the square of DG or DB, is equivalent 

to the square of DA or to the squares of AB 

and DB, the rectangle EA, AG, or FA, AC, is equivalent 

to the square of AB. 

PROBLEM XXXVI. 

Given either one of the sides AB, or the base a b, to construct 
an isosceles triangle, so that each of the angles at the base may 
be double of its vertical angle. 

First, let one of the sides AB be given. c 

By the last problem divide it into two parts, ^ B 

AC, CB, such that CB 2 =:AB x AC. Construct the trian- 
gle, having the base — CB, and each of the two sides =AB. 

Next, if the base AB be given, by the sec- a B C 

ond case of the foregoing proposition, produce 

AB to C, so that ACxCB=AB 2 ; then will AB be the 
base, and AC the length of each of the two sides. 

PROBLEM XXXVH. 

To describe a regular pentagon on a given line AB. 

On AB erect the isosceles triangle 
ACB, having each of the angles at the 
base double of its vertical angle ; on AB 
again construct another isosceles trian- 
gle, whose vertical angle AOB is double 
of ACB, and about the vertex O place 
the isosceles triangles AOD, DOC, 
COE, and EOB ; these triangles, with 
AOB, will compose a regular pentagon. 

For the angle AOB, being the double of ACB, which is 
the fifth part of two right angles, must be equal to the fifth 
part of four right angles; and, consequently, five angles, each 
of them equal to AOB, will adapt themselves about the point 
O. But the bases of those central triangles, and which form 
the sides of the pentagon, are all equal ; and the angles at 
their bases being likewise equal, they are equal in the col- 
lective pairs which constitute the internal angles of the fig- 
ure. It is therefore a regular pentagon. 




w •;•-- 
^;<:: 



PROBLEMS.^ 85 



PROBLEM XXXVTII. 

To describe a hexagon upon a given line AB. ^ 

From A and B as centres, with AB as radius, describe 
arcs intersecting in O (fig. to the next problem). From O 
as a centre, with the same radius, describe a circle ABCDEF. 
Within this circle set off from B the chords BC, CD, DE, 
EF, FA, in succession, each equal to AB ; they will, to- 
gether with AB, form the hexagon required. 

The demonstration is analogous to that of the following 
problem. 

PROBLEM XXXIX. 

To inscribe a regular hexagon in a circle. 

Apply the radius AO of the given ~F^-~r<& 
circle as a chord, AB, BC, CD, &c, /^\ /\\ 

quite round the circumference, and it // \ / \\ 
will complete the regular hexagon A/ __ K W J^S-n 
ABCDEF. vC^^^T^yF 

For, draw the radii AO, BO, CO, \\ / \ \ *// 
DO, EO, FO, completing six equal tri- x/l^-^^ 

angles, of which any one, as ABO, be- B 

ing equilateral (by constr.), its three angles are all equal 
(Cor. 2, th. 3, B. I.), and any one of them, as AOB, is one 
third of the whole, or of two right angles (th. 17, B. I.), or 
one sixth of four right angles. But the whole circumfer- 
ence is the measure of four right angles (Cor. 4, th. 6, B. I.). 
Therefore the arc AB is one sixth of the circumference of 
the circle, and, consequently, its chord AB one side of an 
equilateral hexagon inscribed in the circle. And the same 
of the other chords. 

Cor. The side of a regular hexagon is equal to the radius 
of the circumscribing circle, or to the chord of one sixth 
part of the circumference. 

PROBLEM XL. 

On a given line AB, to construct a regular octagon. 
Bisect AB by the perpendicular CD, which make = CA 



86 



GEOMETRY. 



or CB ; join DA and DB ; produce 
CD, making DO = DA or DB ; draw 
AO and BO, thus forming an angle 
equal to the half of ADB, and about 
the vertex O repeat the equal trian- 
gles AOB, AOE, EOF, FOG, GOH, 
HOI, IOK, and KOB, to compose 
the octagon. 

For the distances AD, BD are evi- 
dently equal ; and because CA, CD, 
and CB are all equal, the angle ADB is contained in a semi- 
circle, and is, therefore, a right angle. Consequently, AOB 
is equal to the half of a right angle, and eight such angles 
will adapt themselves about the point O. Whence the fig- 
ure BAEFGHIK, having eight equal sides and equal angles, 
is a regular octagon. 




PROBLEM XLI. 

To divide the circumference of a given circle successively into 
4, 8, 12, and 24 equal parts. 

1. Insert .the radius AB three times 
from A to D, E, and C ; from the extrem- 
ities of the diameter AC, and with a dis- 
tance equal to the double chord AE, de- 
scribe arcs intersecting in the point F ; 
and from A, with the distance BF, cut the 
circumference on opposite sides at G and 
H ; AG, GC, CH, and HA are quadrants. 

2. From the point F, with the radius 
AB cut the circle in I and K, and from A and C inflect the 
chord AI from L and M ; the circumference is divided into 
eight equal portions by the points A, I, G, K, C, M, H, 
and L. 

3. The arc DG, on being repeated, will form twelve equal 
sections of the circumference. 

4. The arc ID is the twenty-fourth part of the circum- 
ference. 




PROBLEMS. 



87 



PROBLEM XLH. 

To divide the circumference of a given circle successively into 
5, 10, and 20 equal parts. 

Mark out the semicircumference ADEC 
by the triple insertion of the radius ; from 
A and C, with the double chord AE, de- 
scribe arcs intersecting in F ; from A, with 
the distance BF, cut the circle in G- and 
H ; inflect the chords GI and GI equal 
to the radius AB ; and from the points I 
and I, with distance BF or AG, describe 
arcs intersecting in L. 

For BL is the greater segment of the radius BH divided 
by a medial section ; wherefore AL is equal to the side of 
the inscribed pentagon, and BL to that of the decagon in- 
scribed in the given circle. Hence AL may be inflected five 
times in the circumference, and BL ten times ; and, conse- 
quently, the arc MK, or the excess of the fourth above the 
fifth, is equal to the twentieth part of the whole circumfer- 
ence. 




PROBLEM XLIII. 

To describe a regular pentagon, hexagon, or octagon about a 
circle. 

In the given circle inscribe a regular 
polygon of the same name or number of 
sides, as ABCDE, by one of the foregoing 
problems. Then to all its angular points 
draw tangents (by prob. 22), and these will 
form the circumscribing polygon required. 

For all the chords or sides of the in- £ 

scribed figure, AB, BC, &c, being equal, and all the radii 
OA, OB, &c, being equal, all the vertical angles about the 
point O are equal. But the angles OEF, OAF, OAG, 
OBG, made by the tangents and radii, are right angles ; 
therefore OEF+OAF=two right angles, and OAG+OBG 
=two right angles; consequently, also, AOE+AFE=two 
right angles, and AOB-f AGB=two right angles (Cor. 2, 




88 GEOMETRY. 

th. 18, B. I.). Hence, then, the angles AOE + AFE being 
=AOB+AGB, of which AOB is=:AOE, consequently 
the remaining angles F and G- are also equal. In the same 
manner it is shown that all the angles F, G, H, I, K are 
equal. 

Again, the tangents from the same point FE, FA are 
equal, as also the tangents AG, GB (Cor. 2, th. 21, B. II.) ; 
and the angles F and G of the isosceles triangles AFE, 
AGB, are equal, as well as their opposite sides AE, AB ; 
consequently, those two triangles are identical (th. 1), and 
have their other sides EF, FA, AG, GB, all equal, and FG 
equal to the double of any one of them. In like manner it 
is shown that all the other sides GH, HI, IK, KF, are equal 
to FG-, or double of the tangents GB, BH, &c. 

Hence, then, the circumscribed figure is both equilateral 
and equiangular, which was to be shown. 

Cor. The inscribed circle touches the middle of the sides 
of the polygon. 

PROBLEM XLIV. 

To inscribe a circle in a regular polygon. 

Bisect any two sides of the polygon by 
the perpendiculars GO, FO, and their inter- 
section O will be the centre of the inscribed j^ 
circle, and OG or OF will be the radius. 

For the perpendiculars to the tangents 
AF, AG- pass through the centre (Cor., th. 
7, B. H.), and the inscribed circle touches 
the middle points F, G by the last corollary. Also, the two 
sides AG, AO, of the right-angled triangle AOG, being 
equal to the two sides AF, AO of the right-angled triangle 
AOF, the third sides OF, OG will also be equal. There- 
fore the circle described with the centre O and radius OG 
will pass through F, and will touch the sides in the points G 
and F. And the same for all the other sides of the figure. 





PROBLEMS. 89 

PROBLEM XLV. 

To describe a circle about a regular polygon. 

Bisect any two of the angles C and D 
with the lines CO, DO ; then their inter- 
section O will be the centre of the circum- 
scribing circle, and OC or OD will be the j 
radius. 

For, draw OB, OA, OE, &c, to the an- 
gular points of the given polygon. Then C 
the triangle OCD is isosceles, having the angles at C and D 
equal, being the halves of the equal angles of the polygon 
BCD, CDE ; therefore their opposite sides CO, DO are 
equal (th. 4, B. I.). But the two triangles OCD, OCB, 
having the two sides OC, CD equal to the two OC, CB, 
and the included angles OCD, OCB also equal, will be iden- 
tical (th. 1, B. I.), and have their third sides BO, OD equal. 
In like manner it is shown that all the lines OA, OB, OC, 
OD, OE, are equal. Consequently, a circle described with 
the centre O and radius OA will pass through all the other 
angular points B, C, D, &c, and will circumscribe the po- 
lygon. 

PROBLEM XLVT. 

On a given line, to construct a rectilinear figure similar to a 
given rectilinear figure. 

Let abcde be the given 
rectilinear figure, and AB 
the side of the proposed sim- 
ilar figure that is similarly 
posited with ab. 

Place AB in the prolonga- 
tion of ab, or parallel to it. 

Jbw AC, AD, AE, &c, parallel to ac, ad, ae respectively. 
Draw BC parallel to be, meeting AC in C ; CD parallel to 
cd, and meeting AD in D ; DE parallel to de, and meeting 
AE in E, and so on till the figure is completed. Then 
ABCDE will be similar to abcde, from the nature of paral- 
lel lines and similar figures (th. 11, B. IV.). 




90 GEOMETEY. 



PROBLEM XLVII. 



Through a given point within an angle, to draw a straight 
line such that the parts included between the point and the 
of the angle may be 



Let B be the given point, and AED 
the angle. 

It is required to draw a line AD, 
through B, so that AB may be equal 
toBD. 

Through the point B draw BC par- 
allel to DE, and take CA=CE. B 

Through the points A and B draw the line ABD, which 
will be the line required. 

For BC is parallel to DE by construction. 

Therefore AC : CE : : AB : BD (th. 4, B. IV.). 

But AC is equal to CE by construction ; 
/. AB is equal to BD. 




GEOMETRY OT PLANES, 



book y. 

DEFINITIONS. 



1. A plane is a surface in which, if any two points be 
taken, the straight line which joins these points will be 
wholly in that surface. 

2. A straight line is said to be perpendicular to a plane 
when it is perpendicular to all the straight lines in the plane 
which pass through the point in which it meets the plane. 

This point is called the foot of the perpendicular. 

3. The inclination of a straight line to a plane is the 
acute angle contained by the straight line and another 
straight line drawn from the point in which the first meets 
the plane to the point in which a perpendicular to the 
plane, drawn from any point in the first line, meets the 
plane. 

4. A straight line is said to be parallel to a plane when 
it can not meet the plane, to whatever distance both be pro- 
duced. 

5. It will be proved in Prop. 2, that the common inter- 
section of two planes is a straight line ; this being premised, 

The angle contained by two planes which cut one an- 
other is measured by the angle contained by two straight 
lines drawn from any point in the common intersection of 
the planes perpendicular to it, one in each of the planes. 

This angle may be acute, right, or obtuse. 

If it be a right angle, the planes are said to be perpen- 
dicular to each other. 

6. Two planes are parallel to each other when they can 
not meet, to whatever distance both be produced. 



92 



GEOMETRY OF PLANES. 



PROPOSITION I. 



A straight line can not be partly in a plane and partly out 
of it. 

For, by Def. 1, when a straight line has two points com- 
mon to a plane, it lies wholly in that plane. 



A. 


D 












^ 


:> 


B 



PROPOSITION n. 

If two planes cut each other, their common intersection is a 
straight line. 

Let the two planes AB, CD cut one an- 
other, and let P, Q be two points in their 
common section. 

Join P, Q ; 

Then, since the points P, Q are in the 
same plane AB, the straight line PQ which 
joins them must lie wholly in that plane. 

For a similar reason, PQ must lie whol- 
ly in the plane CD. 

.". The straight line PQ is common to 
the two planes, and is .*. their common intersection. 

PROPOSITION I]J. 

Any number of planes may be drawn through the same 
straight line. 

For let a plane, drawn through a straight line, be con- 
ceived to revolve round the straight line as an axis. Then 
the different positions assumed by the revolving plane will 
be those of different planes drawn through the straight line. 

PROPOSITION *EV. 

One plane, and one plane only, can be drawn 

1°. Through a straight line and a point not situated in the 
given line. 

2°. Through three points which are not in the same straight 
line. 

3°. Through two straight lines which intersect each other. 

4°. Through tivo parallel straight lines. 



GEOMETRY OF PLANES. 



93 



1. For if a plane be drawn through the given line, and 
be conceived to revolve round it as an axis, it must, in the 
course of a complete revolution, pass through the given point, 
and so assume the position enounced in 1°. 

Also, one plane only can answer these conditions ; for if 
we suppose a second plane passing through the same straight 
line and point, it must have at least two intersections with 
the first, which is impossible. 

2. Join two of the points : this case is then reduced to the 
last. 

3. Take a point in each of the lines which is not the 
point of intersection : join these two points : the case is now 
the same as the two former. 

4. Parallel straight lines are, by their definition, in the 
same plane, and, by the first case, one plane only can be 
drawn through either of them and a point assumed in the 
other. 

Cor. Hence the position of a plane is determined by 

1. A straight line and a point not in the given straight line. 

2. A triangle, or three points not in the same straight line. 

3. Two straight lines which intersect each other. 

4. Tivo parallel straight lines. 



PROPOSITION V. 

If a straight line he perpendicular to two other straight lines 
which intersect at its foot in a plane, it ivill be perpendicular to 
every other straight line drawn through its foot in the same plane, 
and will therefore be perpendicular to the plane. 

Let XZ be a plane, and let the straight 
line PQ be perpendicular to the two 
straight lines AB, CD, which intersect 
in Q in the plane XZ. 

Draw any straight line EF through 
Q; 

Then PQ will be perpendicular to EF. 

Draw through any point K in QF a 
straight line GH, such that GK=KH. 

JoinP, G; P, K; P, H ; 

Then, since GH, the base of the A GQH, is bisected 
.\ GQ 2 +HQ 2 =2GK 2 +2QK 2 .... (1) 




inK, 



94 



GEOMETRY OP PLANES. 



Similarly, since GH, the base of A GPH, is bisected in K, 

... GP 2 +HP 2 =2GK 2 +2PK 2 . 
But the angles PQG, PQH are right angles ; .*. the above 
becomes 

PQ 2 +GQ 2 +PQ 2 +HQ 2 =2GK 2 4-2PK 2 . . . (2) 
Taking (1) from (2), there remains 

2PQ 2 =2PK 2 -2QK 2 ; 
.-.PQ 2 H-QK 2 =PK 2 . 
Hence PQK is a right angle. 

In like manner, it may be proved that PQ is at right an- 
gles to every other straight line passing through Q in the 
plane XZ. 




PROPOSITION VI. 

A perpendicular is the shortest line which can be drawn 
from any point to a plane. 

Let PQ be perpendicular to the plane 
XZ; 

From P draw any other straight line 
PK to the plane XZ ; 

Then PQ<PK. 

In the plane XZ draw the straight 
line QK, joining the points Q, K. 

Then, since the line PQ is perpendic- 
ular to the plane XZ, the angle PQK is a right angle ; and 
.-. PQ is less than any other line PK (Geom., th. 21, B. I.). 

Cor. 1. Hence oblique lines equally distant from the 
perpendicular are equal ; and, if two oblique lines be une- 
qually distant from the perpendicular, the more distant is 
the larger. 

That is, if QG, QH, QK, are all 

equal, then PG, PH, PK, are all 

equal ; and if QH be greater than QG, 
then PH is greater than PG. 

Cor. 2. A perpendicular measures the 
distance of any point from a plane. The 
distance of one point from another is 
measured by the straight line joining 
them, because this is the shortest line 
which can be drawn from one point to another. So, also, 




GEOMETRY OF PLANES. 95 

the distance from a point to a line is measured by a per- 
pendicular, because this line is the shortest that can be 
drawn from the point to the line. In like manner, the dis- 
tance from a point to a plane must be measured by a per- 
pendicular drawn from the point to the plane, because this 
is the shortest line that can be drawn from the point to the 
plane. 

proposition vn. 

Let PQ be a perpendicular on the plane XZ, and GH a 
straight line in that plane ; if from Q, the foot of the perpen- 
dicular, QK be drawn perpendicular to GH, and P, K be 
joined, then PK will be perpendicular to GH. 

Take KG=KH ; join P, G ; P, H ; 
Q,G; Q,H; 

yKGxzKH, and KQ common to 
the triangles GQK, HQK, and angle 
GKQ= angle HKQ, each being a right 
angle, 

.-.QG=QH; 

.-. PG=PH. Cor. to last Prop. 

Hence the two, triangles GKP, 
HKP have the two sides GK, KIP equal to the two sides 
HK, KP, and the remaining side GP equal to the remain- 
ing side HP. 

/.Angle GKP = angle HKP; and.-, each of them is a 
right angle. 

Cor. GH is perpendicular to the plane PQK, for GH is 
perpendicular to each of the two straight lines KP, KQ. 

Remark. — The two straight lines PQ, GH present an 
example of two straight lines which do not meet, because 
they are not situated in the same plane. 

The shortest distance between these two lines is the 
straight line QK, which is perpendicular to each of them. 

For, join any other two points, as P, G; 

Then, PG>PK") , . ^ 

And, KP>KQ| la4Pt0 P- 5 

Therefore PG>KQ. 

The two lines PQ, GH, although not situated in the 
same plane, are considered to form a right angle with each 




96 GEOMETRY OF PLANES. 

other. For PQ and a straight line drawn through any 
point in PQ parallel to GH would form a right angle. 

In like manner, PG- and QK, which represent any two 
straight lines not situated in the same plane, are considered to 
form with each other the same angle which PG would make 
with any parallel to QK, drawn through a point in PG. 

proposition vin. 

Iftivo straight lines he perpendicular to the same plane, they 
will he parallel to each other. 

Let each of the straight lines PQ, GH 
be perpendicular to the plane XZ. 

Then PQ will be parallel to GH. 

In the plane XZ draw the straight line 
QH, joining the points Q, H. 

Then, since PQ, GH are perpendicular 
to the plane XZ, they are perpendicular to 
the straight line QH in that plane ; and, 
since PQ, GH are both perpendicular to the same line QH, 
they are parallel to each other (Geom., th. 14, Cor., B. I.). 

Cor. 1. Conversely, if two straight lines be parallel, and 
if one of them be perpendicular to any plane, the other will 
also be perpendicular to the same plane. 

Cor. 2. Two straight lines parallel to a third are paral- 
lel to each other. 

For, conceive a plane perpendicular to any one of them, 
then the other two, being parallel to the first, will be per- 
pendicular to the same plane ; hence, by the Prop., they will 
be parallel to each other. 

The three straight lines are not supposed to be in the 
same plane : in this case the proposition has been already 
demonstrated. 

PROPOSITION IX. 

If a straight line without a given plane be parallel to a 
straight line in the plane, it will be parallel to the plane. 

Let AB, lying without the plane XZ, be parallel to CD, 
lying in the plane : 

Then AB is parallel to the plane XZ. 



GEOMETRY OP PLANES. 97 

Through the parallels AB, CD draw a b 

the plane ABCD. I — j 

If the line AB can meet the plane XZ, 
it must meet it in. some point of the line x 

CD, which is the common intersection of [~~ \ 

the two planes. I £ =| V 

But AB can not meet CD, because l_ \ 

AB is parallel to CD. z 

Hence AB can not meet the plane XZ ; i e., AB is paral- 
lel to the plane XZ. 

proposition x. 

The sections made by a plane cutting two parallel planes are 
parallel. 

Let FE, GH be the sections made 
by the plane GF, which cuts the paral- 
lel planes XZ, WY ; 

Then FE will be parallel to GH. 

For if the lines FE, GH, which are 
situated in the same plane, be not par- 
allel, they will meet if produced. There- 
fore the planes XZ, WY, in which 
these lines lie, will meet if produced, H Y 

and .-. can not be parallel, which is contrary to the hypothe- 
sis. 

.-. FE is parallel to GH. . 

PROPOSITION XI. 

Parallel straight lines included between two parallel planes 
are equal. 

Let the parallels EG, FH be cut by 
the parallel planes XZ, WY, in the 
points G, H, E, F ; 

Then EG=rFH. 

Through the parallels EG, FH draw 
the plane EGHF, intersecting the par- 
allel planes in GH, FE. 

Then GH is parallel to FE, by last 
Prop. ; 

E 



E 


X 


1 ,/\ 


/ 


Z FY \ 
W \ 


V 


\ V* 


^"\ 





98 GEOMETRY OF PLANES. 

And GE is parallel to HF ; 

/. GHFE is a parallelogram ; and therefore 

EG^FH. 
Cor. Two parallel planes are every where equidistant. 

PROPOSITION XII. 

If two -planes he parallel to each other, a straight line which 
is perpendicular to one of the planes will he perpendicular to the 
other also. 

Let the two planes XZ, WY be par- 
allel, and let the straight line AB be 
perpendicular to the plane XZ; 

Then AB will be perpendicular to 
WY. 

For, from any point H in the plane 
WY draw HG perpendicular to the 
plane XZ, and draw AG, BH. 

Then, since BA, HG are both perpendicular to XZ, /. 
the angles A, G are right angles. 

And, since the planes XZ, "v\Y are parallel, /. the per- 
pendiculars BA, HG are equal. 

Hence AG is parallel to BH, and AB, being perpendicu- 
lar to AG, is perpendicular to BH also. 

In like manner, it may be proved that AB is perpendicu- 
lar to all other lines which can be drawn from B in the 
plane WY. 

.*. AB is perpendicular to the plane WY. 

Cor. Conversely, if two planes be perpendicular to the 
same straight line, they will be parallel to each other. 

proposition xin. 

If two straight lines which form an angle he parallel to two 
other straight lines ivhich form an angle in the same direction, 
although not in the same plane ivith the former, the two angles 
will he equal, and their planes will he parallel. 

Let the two straight lines AB, BC, in the plane XZ, be 
parallel to the two DE, EF, in the plane WY ; 
Then angle ABC=angle DEF. 



GEOMETRY OF PLANES. 



99 




For, make AB=DE, BC=EF 
join A, C ; D, F ; A, D ; B, E; C, F 

Then the straight lines AD, BE, 
which join the equal and parallel 
straight lines AB, EJE, are themselves 
equal and parallel. 

For the same reason, CF, BE are 
equal and parallel. 

.-. AD, CF are equal and parallel, 
and /. AC, DF are also equal and par- 
allel. 

Hence the two triangles ABC, DEF, having all their sides 
equal each to each, have their angles also equal. 
.-. angle ABC = angle DEF. 

Again, the plane XZ is parallel to the plane WY. 

For if not, let a plane drawn through A, parallel to DEF, 
meet the straight lines FC, EB, in G- and H. 

Then D A = EH = FG, Prop. 

But DA=EB=FC. 

Therefore EH =EB, FG=FC, 

which is absurd ; hence, 

Cor. 1. If two parallel planes XZ, 
WY, are met by two other planes 
ADEB, CFEB, the angles ABC, 
DEF, formed by the intersection of 
the parallel planes, will be equal. 

For the section AB is parallel to 
the section DE, Prop. 

So also the section BC is parallel 
to the section EF. 

.-. angle ABC = angle DEF. 
. Cor. 2. If three straight lines AD, BE, CF, not situated 
in the same plane, be equal and parallel, the triangles ABC, 
DEF, formed by joining the extremities of these straight 
lines, will be equal, and their planes will be parallel. 




100 



GEOMETRY OF PLANES. 



PROPOSITION XIV. 

If two straight lines be cut by parallel planes, they will be 
cut in the same ratio. 

Let the straight lines AB, CD be cut 
by the parallel planes XZ, WY, VS, 
in the points A, E, B, C, F, D ; 

Then, AE : EB::CF : FD. 

Join A, C ; B, D ; A, D ; and let 
AD meet the plane WY in G; join 
E, G ; G, F ; 

Then the intersections EG, BD, of 
the parallel planes WY, VS, are paral- 
lel (Prop. 10). 

.-.AE :EB::AG: GD. 

Again, the intersections AC, GF, of the parallel planes 
XZ, YW, with the plane CG, are parallel ; 
/.AG: GD::CF: FD. 

/. comparing this with the first proportion, 
AE: EB::CF:FD. 




PROPOSITION XV. 

If a straight line be at right angles to a plane, every plane 
which passes through it will be at right angles to that plane. 

Let* the straight line PQ be at right an- 
gles to the plane XZ. 

Through PQ draw any plane PO, in- 
tersecting XZ in the line OQW. 

Then the plane PO is perpendicular to 
the plane XZ. 

Draw PS in the plane XZ perpendicu- 
lar to WQO. 

Then, since the straight line PQ is per- 
pendicular to the plane XZ, it is perpen- 
dicular to the two straight lines PS, OW, which pass 
through its foot in that plane. 

But the angle PQR, contained between PQ,, QR, which 
are perpendiculars to OW, the common intersection of the 
planes XZ, PO, measures the angle of the two planes (def. 



*"!£-* 


PT 




\ 


X 


s \ 






1Y* 


V 



GEOMETRY OF PLANES. 



101 



5) ; hence, since this angle is a right angle, the two planes 
are perpendicular to each other. 

Cor. If three straight lines, such as PQ, RS, OW, be per- 
pendicular to each other, each will be perpendicular to the 
plane of the two others, and the three planes will be per- 
pendicular to each other. 



\ 



w 



\ 



PROPOSITION XVI. 

If two planes be perpendicular to each other, a straight line 
drawn in one of the planes perpendicular to their common sec- 
tion will be perpendicular to the other plane. 

Let the plane VO be perpendicular to 
the plane XZ, and let OW be their com- 
mon section. 

In the plane VO draw PQ perpendicu- 
lar to OW ; 

Then PQ is perpendicular to the plane 
XZ. 

From the point Q draw QR. in the 
plane XZ perpendicular to OW. 

Then, since the two planes are perpen- Z — 
dicular, the angle PQR is a right angle. 

.*. The straight line PQ is perpendicular to the straight 
lines QR, QO, which intersect at its foot in the plane XZ. 

.*. PQ is perpendicular to the plane XZ. 

Cor. If the plane VO be perpendicular to the plane XZ, 
and if from any point in OW, their common intersection, 
we erect a perpendicular to the plane XZ, that straight line 
will lie in the plane VO. 

For if not, then we may draw from the same point a 
straight line in the plane VO perpendicular to OW, and this 
line, by the Prop., will be perpendicular to the plane XZ. 

Thus we should have two straight lines drawn from the 
same point in the plane XZ, each of them perpendicular to 
the given plane, which is absurd. 



102 



GEOMETKY OF PLANES. 



PROPOSITION XVII. 

If two planes which cut each other he each of them perpen- 
dicular to a third plane, their common section will be perpendic- 
ular to the same plane. 

Let the two planes VO, TW, whose 
common intersection is PQ, be both per- 
pendicular to the plane XZ. 

Then PQ is perpendicular to the plane 
XZ. 

For, from the point Q erect a perpen- 
dicular to the plane XZ. 

Then, by the Cor. to the last Prop., this 
straight line must be situated at once in 
the planes VO and TW, and is therefore 
their common section. 



T 


X 







SOLID ANGLES, 



BOOK YI. 

DEFINITIONS. 



1. A solid angle is the angular space contained between 
several planes which meet in the same point. 

2. Three planes, at least, are required to form a solid angle. 

3. A solid angle is called a trihedral, tetrahedral, &c, an- 
gle, according as it is formed by three, four, . . . .plane angles. 

4. If three planes intersect each other in a common point, 
the indefinite space bounded by those -three planes is called 
a triangular pyramidal space. 

5. If three planes intersect each other in parallel lines, 
the indefinite space bounded by those three planes is called 
a triangular prismatic space. 

6. If four or more planes intersect each other in a com- 
mon point, the indefinite space bounded by those planes is 
called a polyhedral pyramidal space. 

7. If four or more planes intersect each other in parallel 
edges, the indefinite space bounded by them is called a poly- 
hedral prismatic space. 

PROPOSITION I. 

If a solid angle be contained by three plane angles, the sum 
of any two of these angles will be greater than the third. 

It is unnecessary to demonstrate this 
proposition except in the case where the 
plane angle which is compared with the 
two others is greater than either of them. 

Let A be a solid angle contained by 
the three plane angles BAG, CAD, DAB, 
and let BAG be the greatest of these an- 
gles; 




104 



GEOMETRY. 



Then, CAD+DAB>BAC. 

In the plane BAC draw the straight line AE, making the 
angle BAE = angle BAD. 

Make AE=:AD, and through E draw any straight line 
BEC, cutting AB, AC, in the points B, C ; join D, B ; D, C ; 
Then, v AD=AE, and AB is common to the two trian- 
gles DAB, BAE, and the angle DAB— angle BAE, 
.-.BD = BE. 
But BD+DC>BE + EC, 

/.DC>EC 
Again, v AD=:AE, and AC is common to the two tri- 
angles DAC, EAC, but the base DC>base EC, 
.-. angle DAC>augle EAC. 
But angle DAB = angle BAE. 

.-. angle CAD + angle DAB > angle BAE + angle EAC 

> angle BAC. 



proposition n. 

The sum of the plane angles which form a solid angle is 
always less titan four right angles. 

Let P be a solid angle contained by any 
number of plane angles APB, BBC, CPD, 
DPE, EPA. 

Let the solid angle P be cut by any plane 
ABCDE. 

Take any point, O, in this plane; join 
A, O; B, O; C,0; D,0; E,0; 

Then, since the sum of all the angles of 
every triangle is always equal to two right 
angles, the sum of all the angles of the triangles ABP, BBC, 
.... about the point P, will be equal to the sum of all the 

angles of the equal number of triangles AOB, BOC, 

about the point O- 

Again, by the last proposition, angle ABC < angle ABP 
+ angle CBP; in like manner, angle BCD<angle BCP 
-j- angle DCP, and so for all the angles of the polygon 
ABCDE. 

Hence the sum of the angles at the bases of the triangles 
whose vertex is O is less than the sum of the angles at the 
bases of the triangles whose vertex is P. 





PROPOSITIONS. 105 

/. The sum of the angles about the point O must be 
greater than the sum of the angles about the point P. 

But the sum of the angles about the point is four right 
angles. 

.*. The sum of the angles about the point P is less than 
four right angles. 

proposition m. 

If two solid angles be formed by three plane angles which 
are equal, each to each, the planes in which these angles lie will 
be equally inclined to each other. 

Let P, Q, be two solid angles, 
each contained by three plane an- 
gles ; 

Let angle APC = angle DQF, 
angle APB wangle DQE, and an- a/ 
gle BPC= angle EQF. 

Then, the inclination of the 
planes APC, APB, will be equal to the inclination of the 
planes DQF, DQE. 

Take any point, B, in the intersection of the planes APB, 
CPB. 

From B draw BY perpendicular to the plane APC, meet- 
ing the plane in Y. 

From Y draw YA, YC, perpendiculars on PA, PC ; join 
A, B ; B, C ; 

Again, take QE=rPB, from E draw EZ perpendicular to 
the plane DQF, meeting the plane in Z; from Z draw ZD, 
ZF, perpendiculars on QD, QF ; join D, E ; E, F. 

The triangle PAB is right angled at A, and the triangle 
QDE is right angled at D. (Geom. of Planes, Prop. 7.) 

Also, the angle APB wangle DQE, by construction. 
.v angle PBA= angle QED. 

But the side PB=side QE, .-. the two triangles APB, 
DQF, are equal and similar. 

.-. PA = QD, and AB =DE. 

In like manner, we can prove that 

PC=QF, and BC=EF. 

We can now prove that the quadrilateral PAYC is equal 
to the quadrilateral QDZF. 

E 2 ' 



106 GEOMETRY. 

For, let the angle APC be placed upon the equal angle 
DQF, then the point A will fall upon the point D, and the 
point C on the point F, because PA=QD, and PC=:QF. 

At the same time, AY, which is perpendicular to PA, 
will fall upon DZ, which is perpendicular to QD ; and, in 
like manner, CY will fall upon FZ. 

Hence the point Y will fall on the point Z, and we shall 
have' 

AY=r:DZ, and CY=FZ. 

But the triangles AYB, DZE, are right angled at Y and 
Z, the hypothenuse AB= hypothenuse DE, and the side AY 
reside DZ; hence these two triangles are equal. 
.-. angle YAB = angle ZDE. 

The angle YAB is the inclination of the planes APC, 
APB; and 

The angle ZDE is the inclination of the planes DQF, DQE. 
.*. these planes are equally inclined to each other. 

In the same manner, we prove that angle YCB= angle 
ZFE, and, consequently, the inclination of the planes APC, 
BPC, is equal to the inclination of the planes DQF, EQF. 

We must, however, observe that the angle A of the right- 
angled triangle YAB is not, properly speaking, the inclina- 
tion of the two planes APC, APB, except when the per- 
pendicular BY falls upon the same side of PA as PC does ; 
if it fall upon the other side, then the angle between the 
two planes will be obtuse, and, added to the angle A of the 
triangle YAB, will make up two right angles. But, in this 
case, the angle between the two planes DQF, DQE, will 
also be obtuse, and, added to the angle D of the triangle 
ZDE, will make up two right angles. 

Since, then, the angle A will always be equal to the an- 
gle D, we infer that the inclination of the two planes APC, 
APB, will always be equal to the inclination of the two 
planes DQF, DQE. In the first case, the inclination of the 
plane is the angle A or D ; in the second case, it is the sup- 
plement of those angles. 

Scholium. — If two solid trihedral angles have the three 
plane angles of the one equal to the three plane angles of 
the other, each to each, and, at the same time, the corre- 
sponding angles arranged in the same manner in the two sol- 



PROPOSITIONS. 107 

id angles, then these two solid angles will be equal ; and if 
placed one upon the other, they will coincide. In fact, we 
have already seen that the quadrilateral PAYC will coin- 
cide with the quadrilateral QDZF. Thus the point Y falls 
upon the point Z, and, in consequence of the equality of the 
triangles AYB, DZE, the straight line YB, perpendicular to 
the plane APC, is equal to the straight line ZE, perpendic- 
ular to the plane DQE ; moreover, these perpendiculars lie 
in the same direction ; hence the point B will fall upon the 
point E, the straight line PB on the straight line QE, and 
the two solid angles will entirely coincide with each other. 

This coincidence, however, can not take place except we 
suppose the equal plane angles to be arranged in the same 
manner in the two solid angles ; for, if the equal planes be 
arranged in an inverse order, or, which comes to the same 
thing, if the perpendiculars YB, ZE, instead of being situ- 
ated both on the same side of the planes APC, DQF, were 
situated on opposite sides of these planes, then it would be 
impossible to make the two solid angles coincide with each 
other. It would not, however, be less true, according to the 
above theorem, that the planes in which the equal angles lie 
would be equally inclined to each other; so that the two 
solid angles would be equal in all their constituent parts, 
without admitting of superposition. This species of equal- 
ity, which is, not absolute, or equality of coincidence, has re- 
ceived from Legendre a particular description. He terms 
it equality of symmetry. 

Thus the two solid trihedral angles in question, which 
have the three plane angles of the one equal to the three 
plane angles of the other, each to each, but arranged in an 
inverse order, are termed angles equal by symmetry, or simply 
symmetrical angles. 

The same observation applies to solid angles formed by 
more than three plane angles. Thus a solid angle formed 
by the plane angles A, B, C, D, E, and another solid angle 
formed by the same angles in an inverse order, A, E, D, C, B, 
may be such that the planes in which the equal angles are 
situated are equally inclined to each other. These two sol- 
id angles, which would in this case be equal, although not 
admitting of superposition, would be termed solid angles 
equal by symmetry, or symmetrical solid angles. 



108 GEOMETRY. 

In plane figures there is no species of equality to which 
this designation can belong, for all those cases to which the 
term might seem to apply are cases of absolute equality, or 
equality of coincidence. The reason of this is that the posi- 
tion of a plane figure may be altered at pleasure, and one 
may take the upper part for the under, and vice versa. This, 
however, does not hold in solids, in which the third dimen- 
sion may be taken in two different directions. 

PROPOSITION IV. 

If three planes intersect each other in three lines, these lines 
ivill either meet in the same common point, or they will be par- 
allel to each other. 

Let AE, FC, and BD be three 
planes intersecting each other re- 
spectively in the lines EF, AB, 
and DC ; then each pair of these 
lines will be in the same plane. 
Now each pair of these lines must 
be either parallel or must inter- 
sect in a common point. (Prop. 4.) 

1°. If two of these lines of in- 
tersection, as AB and CB, meet in a point B, that point 
is in the line AB, and also in the planes ADB and DCB ; 
but the point B is, at the same time, a point of the line 
CB, and also in both planes ADB, ACB. The point B is 
therefore in all the three planes, ADB, DBC, ABC, and 
is also one of the points common to the two planes BDC, 
ACB, being a point of the line DB. This third line, DB, 
therefore, meets the same point, B, as the lines AB, CB 
do ; that is, all these lines meet in the same point. 

2°. But if the two lines of intersection, AB, FE, be par-, 
allej. to each other, the third line, DC, can never meet ei- 
ther of these lines, because, if it can meet either of them, 
then the other must also intersect at the same point (Dem. 1), 
and the first two not parallel, which is contrary to the sup- 
position. 

Again, as every two of these lines of intersection are lines 
of the same plane, and can not meet, therefore they must all 
be parallel to each other. 




SOLID GEOMETRY, 

BOOK YII. 

DEFINITIONS. 

1. Similar solid figures are such as have all their solid 
angles equal, each to each, and are contained by the same 
number of similar planes. 

2. A pyramid is a solid figure contained by planes that 
are constituted between one plane and one point above it, 
in which they meet. 

3. A prism is a solid figure contained by plane figures, 
of which two that are opposite are equal, similar, and par- 
allel to each other, and the others are parallelograms. 

4. A sphere is a solid figure described by the revolution 
of a semicircle about its diameter, which remains un- 
moved. 

Thus the inner side of the semicircle revolving round the 
diameter, which remains fixed, generates a sphere. . 

5. The axis of a sphere is the fixed right line about which 
the semicircle revolves. 

6. The centre of a sphere is the same with that of the 
semicircle. 

7. The diameter of a sphere is any right line which passes 
through the centre, and is terminated both ways by the su- 
perficies of the sphere. 

8. A right cone is a solid figure described by the revolu- 
tion of a right-angled triangle about one of the sides con- 
taining the right angle, which side remains fixed. 

If the fixed side be equal to the other side containing the 
right angle, the cone is called a right-angled cone ; if it be 
less than the other side, an obtuse-angled ; and if greater, 
an acute-angled cone. 



110 



SOLID GEOMETRY. 




Thus the side AC, revolving round AB, 
one of the sides which contains the right 
angle and remains fixed, generates a cone. 



9. The axis of a cone is the fixed right line about which 
the triangle revolves* 

In the preceding figure, AB is the axis. 

10. The base of a cone is the circle described by that side 
containing the right angle which revolves. 

11. A cylinder is a solid figure described by the revolu- 
tion of a right-angled parallelogram about one of its sides, 
which remains fixed. — ., 

^^ 

Thus the revolution of the parallelogram AC 
about its side AB, which remains fixed, gener- 
ates a cylinder. 



12. The axis of a cylinder is the fixed right line about 
which the parallelogram revolves. 

13. The bases of a cylinder are the circles described by 
the two revolving opposite sides of the parallelogram. 

14. Similar cones and cylinders are those which have 
their axes and the diameters of their bases proportionals. 



15. A cube is a solid figure contained by six 
equal squares. 



16. A tetrahedron is a solid figure con- 
tained by four equal and equilateral trian- 
gles. 

17. An octahedron is a solid figure con- 
tained by eight equal and equilateral trian- 
gles. 





SOLID GEOMETRY. Ill 

18. A dodecahedron is a solid figure contained by twelve 
equal pentagons which are equilateral and equiangular. 

19. An icosahedron is a solid figure con- 
tained by twenty equal and equilateral trian- 
gles. 



20. A parallelopiped is a solid figure contained 
by six quadrilateral figures, whereof every opposite 
two are parallel. 



PKOPOSITIONS. 

PROPOSITION I. 

If a prism be cut by a plane parallel to its base, the section 
ioill be equal to and like the base. 

Let AGr be any prism, and IL a plane parallel 
to the base AC, then will the plane IL be equal E 
to and like the base AC, or the two planes will 
have all their sides and all their angles equal. 

For the two planes, AC, IL, being parallel, 
by hypothesis ; and two parallel planes, cut by a 
third plane, having parallel sections; therefore 
IK is parallel to AB, KL to BC, LM to CD, and IM to AD. 
But AI and BK are parallels, by Def. 3 ; consequently, AK 
is a parallelogram, and the opposite sides, AB, IK, are equal. 
In like manner, it is shown that KL is=BC, and LM= 
CD, and IM= AD, or the two planes, AC, IL, are mutu- 
ally equilateral. But these two planes, having their cor- 
responding sides parallel, have the angles contained by them 
also equal ; namely, the angle A=the angle I, the angle B 
= the angle K, the angle C = the angle L, and the angle D 
=the angle M. So that the two planes, AC, IL, have all 
their corresponding sides and angles equal, or are equal and 
like. Q. E. D. 



/ \ 


M 

/ \ 



112 



SOLID GEOMETRY. 



PROPOSITION II. 

If a cylinder be cut by a plane parallel to its base, the section 
will be a circle equal to the base. 

Let AF be a cylinder, and GHI any section 
parallel to the base ABC ; then will GHI be a D 
circle equal to ABC. 

For, let the planes KE, KF, pass through G 
the axis of the cylinder MK, and meet the sec- 
tion GHI in the three points H, I, L, and join 
the points as in the figure. 

Then, since KL, CI, are parallel, and the plane 
KI, meeting the two parallel planes ABC, GHI, makes the 
two sections KC, LI, parallel ; the figure KLIC is therefore 
a parallelogram, and, consequently, has the opposite sides 
LI, KC equal, where KC is a radius of the circular base. 

In like manner, it is shown that LH is equal to the ra- 
dius KB ; and that any other lines, drawn from the point 
L to the circumference of the section GHI, are all equal to 
radii of the base ; consequently, GHI is a circle, and equal 
to ABC. Q. E. D. 




proposition ni. 

All prisms, and a cylinder of equal bases and altitudes, are 
equal to each other. 

Let AC, DF, be two 
prisms, and a cylinder upon 
equal bases AB, DE, and 
having equal altitudes ; then 
will the solids AC, DF, be p 
equal. A 

For, let PQ, ES, be any 
two sections parallel to the bases, and equidistant from them. 
Then, by the last two propositions, the section PQ is equal 
to the base AB, and the section ES equal to the base DE. 
But the bases AB, DE, are equal by the hypothesis ; there- 
fore the sections PQ, RS, are also equal. And, in like man- 
ner, it may be shown that any other corresponding sections 
are equal to one another. 



/ 




/ 










/ 






' 




/ 




SOLID GEOMETRY. 



113 



Since, then, every section in the prism AC is equal to its 
corresponding section in the prism, or cylinder PS, the 
prisms and cylinders themselves, which are composed of 
those sections, must also be equal. Q. E. D. 

Corol. Every prism or cylinder is equal to a rectangular 
parallelopipedon of an equal base and altitude. 



9 B g C 



3\ 



^ 



Y 



A L M H E p 



PROPOSITION IV. 

Rectangular parallelopipedons of equal altitudes have to each, 
other the same proportion as their bases. 

Let AC, EG, be two rectangu- 
lar parallelopipedons, having the 
equal altitudes AD, EH; then 
will AC be to EG as the base AB 
is to the base EF. 

For, let the proportion of the 
base AB to the base EF be that 
of any one number m (3) to any other number n (2) ; and 
conceive AB to be divided into m equal parts or rectangles, 
AI, LK, MB (by dividing AJN into that number of equal 
parts, and drawing IL, KM, parallel to BN). And let EF 
be divided, in like manner, into n equal parts or rectangles, 
EO, PF : all of these parts of both bases being mutually 
equal among themselves. And through the lines of divis- 
ion let the plane sections LP, MS, PV pass parallel to 
AQ, ET. 

Then the parallelopipedons AP, LS, MC, EV, PG, are 
all equal, having equal bases and heights. Therefore the 
solid AC is to the solid EG as the number of parts in AC 
to the number of equal parts in EG, or as the number of 
parts in AB to the number of equal parts in EF ; that is, 
as the base AB to the base EF. Q. E. D. 

Corol. From this proposition, and the corollary to the 
la&tg j£ appears that all prisms and cylinders of equal alti- 
tudes are to each other as their bases, every prism and cyl- 
inder being equal to a rectangular parallelopipedon of an 
equal base and height. 



114 SOLID GEOMETRY. 



PROPOSITION V. 



/A 



H 

LZ7 



Rectangular parallelopipedons of equal bases are in propor- 
tion to each other as their altitudes. 

Let AB, CD, be two rectan- 
gular parallelopipedons stand- 
ing on the equal bases AE, CF ; 
then will AB be to CD as the 
altitude EB is to the altitude 
DF. 

For, let AG be a rectangular J£- — ^-^ 
parallelopipedon on the base AE, 

and its altitude EG- equal to the altitude FD of the solid 
CD. 

Then AG and CD are equal, being prisms of equal bases 
and altitudes. But if HB, HG, be considered as bases, the 
solids AB, AG, of equal altitude AH, will be to each other 
as those bases HB, HG. But these bases HB, HG, being 
parallelograms of equal altitude HE, are to each other as 
their bases EB, EG ; and therefore the two prisms AB, AG, 
are to each other as the lines EB, EG. But AG is equal 
CD, and EG equal FD ; consequently, the prisms AB, 
CD, are to each other as their altitudes EB, FD ; that is, 
AB:CD::EB:FD. Q. E. D. 

Cor 61. 1. From this proposition, and the corollary to 
Prop. 3, it appears that all prisms and cylinders of equal 
bases are to one another as their altitudes. 

Corol. 2. Because, by Corol. 1, prisms and cylinders are 
as their altitudes when their bases are equal ; and, by the 
corollary to the last theorem, they are as their bases when 
their altitudes are equal ; therefore, universally, when nei- 
ther are equal, they are to one another as the product of 
their bases and altitudes. And hence, also, these products 
are the proper numeral measures of their quantities or mag- 
nitudes. 

PROPOSITION VI. 

Similar prisms and cylinders are to each other as the cubes 
of their altitudes, or of any other like linear dimensions. 

Let ABCD, EFGH, be two similar prisms; then will 



SOLID GEOMETRY. 



115 



D 


< 


/N 


\ 


















H 


P 1 






/X 












A 


< 


_K 


C E 


t 


7 



the prism CD be to the prism GH 
as AB 3 to EF 3 , or as AD 3 to EH 3 . 

For the solids are to each other as 
the product of their bases and alti- 
tudes (Prop. 5, Cor. 2) ; that is, as AC . 
AD to EG . EH. But the ba^es, be- 
ing similar planes, are to each other 
as the squares of their like sides; that B 
is, AC to EG as AB 2 to EF 2 ; therefore the solid CD is to 
the solid GH as AB 2 . AD to EF 2 . EH. But BD and FH, 
being similar planes, have their like sides proportional ; that 
is, AB : EF : : AD : EH, or AB 2 : EF 2 : : AD 2 : EH 2 ; there- 
fore AB 2 . AD : EF 2 . EH : : AB 3 : EF 3 , or : : AD 3 : EH 3 ; and, 
consequently, the solid CD : solid GH :: AB 3 : EF 3 :: AD 3 
: EH 3 . Q. E. D. 




PROPOSITION vn. 

In a pyramid, a section parallel to the base is similar to the 
base, and these two planes will be to each other as the squares 
of their distances from the vertex. 

Let ABCD be a pyramid, and EFG a sec- 
tion parallel to the base BCD ; also AIH a line 
perpendicular to the two planes at H and I; 
then will BD, EG, be two similar planes, and 
the plane BD will be to the plane EG as AH 2 
toAI 2 . 

For, join CH, FI. Then, because a plane 
cutting two parallel planes makes parallel sec- B 
tions, therefore the plane ABC, meeting the two parallel 
planes BD, EG, makes the sections BC, EF, parallel ; in 
like manner, the plane ACD makes the sections CD, FG, 
parallel. Again, because two pair of parallel lines make 
equal angles, the two EF, FG, which are parallel to BC, 
CD, make the angle EFG equal the angle BCD. And, in 
like manner, it is shown that each angle in the plane EG 
is equal to each angle in the plane BD, and, consequently, 
those two planes are equiangular. 

Again, the three lines AB, AC, AD, making with the 
parallels BC, EF, and CD, FG, equal angles ; and the an- 
gles at A being common, the two triangles ABC, AEF, are 



116 SOLID GEOMETRY. 

equiangular, as also the two triangles ACD, AFG, and 
have therefore their like sides proportional, namely, AC : 
AF : : BC : EF : : CD : FG. And, in like manner, it may be 
shown that all the lines in the plane EG are proportional 
to all the corresponding ones in the base BD. Hence these 
two planes, having their angles equal and their sides pro- 
portional, are similar. 

But similar planes being to each other as the squares of 
their like sides, the plane BD : EG : : BC 2 : EF 2 : or : : AC 2 
: AF 2 , by what is shown above. But the two triangles 
AHC, AIF, having the angles H and I right ones, and the 
angle A common, are equiangular, and have therefore their 
like sides proportional, namely, AC : AF : : AH : AI, or 
AC 2 : AF 2 : : AH 2 : AI 2 . Consequently, the two planes BD, 
EG, which are as the former squares AC 2 , AF 2 , will be 
also as the latter squares AH 2 , AI 2 ; that is, BD : EG : : 
AH 2 :AI 2 . 

PROPOSITION VIII. 

In a right cone, a section parallel to the hase is a circle ; and 
this section is to the base as the squares of their distances from 
the vertex. 

Let ABCD be a right cone, and GHI a 
section parallel to the base BCD ; then will 
GHI be a circle, and BCD, GHI, will be to 
each other as the squares of their distances 
from the vertex. 

For, draw AKE perpendicular to the two 
parallel planes ; and let the planes ACE, ADE, 
pass through the axis of the cone AKE, meet- 
ing the section in the three points H, I, K. 

Then, since the section GHI is parallel to 
the base BCD, and the planes CK, DK, meet them, HK is 
parallel to CE, and IK to DE. And, because the trian- 
gles formed by these lines are equiangular, KH : EC : : AK 
rAE::KI:ED. But EC is equal to ED, being radii of 
the same circle ; therefore KI is also equal to KH. And 
the same may be shown of any other lines drawn from, the 
point K to the circumference of the section GHI, which is 
therefore a circle. 




SOLID GEOMETRY. 



117 



Again, by similar triangles, AK : AE :: KH : EC ; hence 
AK 2 : AE 2 : : KH 2 : EC 2 ; but KH 2 : EC 2 : : circle GHI : cir- 
cle BCD ; therefore AK 2 : AE 2 : : circle GHI : circle BCD. 
Q. E. D. 




PROPOSITION IX. 

All pyramids and right cones of equal bases and altitudes are 
equal to one another. 

Let ABC, DEF, be 
any pyramids and cone 
of equal bases BC, EF, 
and equal altitudes AG, 
DH; then will the pyr- 
amids and cone ABC 
and DEF be equal. 

For, parallel to the 
bases, and at equal distances, AN", DO, from the vertices, 
suppose the planes IK, LM, to be drawn. . 

Then, by propositions 7 and 8, 

D0 2 :DH 2 ::LM:EF, and 
AN 2 : AG 2 :: IK :BC 

But, since AN 2 , AG 2 , are equal to DO 2 , DH 2 ; therefore 
IK : BC : : LM : EF. But BC is equal to EF, by hypothe- 
sis ; therefore IK is also equal to LM. 

In the same manner, it is shown that any other sec- 
tions, at equal distance from the vertex, are equal to each 
other. 

Since, then, every section in the cone is equal to the cor- 
responding section in the pyramids, and the heights are 
equal, the solids ABC, DEF, composed of those sections, 
must be equal also. Q. E. D. 



proposition x. 

Every pyramid of a triangular base is the third part of a 
prism of the same base and altitude. 

Let ABCDEF be a prism, and BDEF a pyramid, upon 
the same triangular base DEF ; then will the pyramid 
BDEF be a third part of the prism ABCDEF. 

For, in the planes of the three sides of the prism, draw 




118 SOLID GEOMETRY. 

the diagonals BF, BD, CD. Then the two planes 
BDF, BCD, divide the whole prism into the B 
three pyramids BDEF, DABC, DBCF, which 
are proved to be all equal to one another as fol- 
lows: 

Since the opposite ends of the prism are equal 
to each other, the pyramid whose base is ABC 
and vertex D is equal to the pyramid whose base is DEF 
and vertex B (Prop. 9), being pyramids of equal base and 
altitude. 

But the latter pyramid, whose base is DEF and vertex 
B, is the same solid as the pyramid whose base is BEF and 
vertex D, and this is equal to the third pyramid, whose 
base is BCF and vertex D, being pyramids of the same al- 
titude and equal bases BEF, BCF. 

Consequently, all the three pyramids which compose the 
prism are equal to each other, and each pyramid is the third 
part of the prism, or the prism is triple of the pyramid. 
Q. E. D. 

Corol. 1. Every pyramid, whatever its figure may be, is 
the third part of a prism of the same base and altitude ; 
since the base of the prism, whatever be its figure, may be 
divided into triangles, and the whole* solid into triangular 
prisms and pyramids. 

Corol. 2. Any right cone is the third part of a cylinder 
or of a prism of equal base and altitude, since it has been 
proved that a cylinder is equal to a prism, and a cone equal 
to a pyramid of equal base and altitude. 

Corol. 3. Every triangular prism may be divided into 
three equal triangular pyramids of the same base and alti- 
tude with the prism. 

Scholium. — Whatever has been demonstrated of the pro- 
portionality of prisms or cylinders holds equally true of 
pyramids or cones, the former being always triple the lat- 
ter, viz., that similar pyramids or cones are as the cubes 
of their like linear sides, or diameters, or altitudes, &Q. 



SOLID GEOMETRY. 119 



PROPOSITION XI. 

If a sphere be cut by a plane, the section will be a circle. 

Because the radii of the sphere are all equal, each of them 
being equal to the radius of the describing semicircle, it is 
evident that if the section pass through the centre of the 
sphere, then the distance from the centre to every point in 
the periphery of that section will be equal to the radius of 
the sphere, and the section will therefore be a circle of the 
same radius as the sphere. But if the plane do not pass 
through the centre, draw a perpendicular to it from the 
centre, and draw any number of radii of the sphere to the 
intersection of its surface with the plane ; then these radii 
are evidently the hypothenuses of a corresponding number 
of right-angled triangles, which have the perpendicular from 
the centre on the plane of the section as a common side ; 
c msequently, their other sides are all equal, and therefore 
.he section of the sphere by the plane is a circle, whose cen- 
tre is the point in which the perpendicular cuts the plane. 

Corol. If two spheres intersect one another, the common 
section is a circle. 

Scholium. — All the sections through the centre are equal 
to one another, and are greater than any other section which 
does not pass through the centre. Sections through the cen- 
tre are called great circles, and the other sections small or less 
circles. Also a straight line drawn through the centre of a 
circle of the sphere perpendicular to the plane of the circle 
is a diameter of the sphere, and the extremities of this di- 
ameter are called the poles of the circle. Hence it is evi- 
dent that the arcs of great circles between the pole and cir- 
cumference are equal, for the chords drawn in the sphere 
from either pole of a circle to the circumference are all 
equal. 

proposition xn. 
Every sphere is two thirds of its circumscribing cylinder. 

Let ABCD be a cylinder circumscribing the sphere 
EFGH ; then will the sphere EFGH be two thirds of the 



x 


"X 1 


j? q\ K| /M^H\ 


V 


y 



120 SOLID GEOMETRY. 

cylinder ABCD. For, let the plane AC 
7 3 a section of the sphere and cylinder 
trough the centre I, and j oin AI, BI. Let 
FIH be parallel to AD or BC, and EIG 
and KL parallel to AB or DC, the base 
of the cylinder ; the latter line KL meet- 
ing BI in M, and the circular section of 
the sphere in N. 

Then, if the whole plane HFBC be conceived to revolve 
about the line HF as an axis, the square FG will describe 
a cylinder AG, and the quadrant IFG will describe a hem- 
isphere EFG, and the triangle IFB will describe a cone 
IAB. Also, in the rotation, the three lines or parts, KL, 
KN, KM, as radii, will describe corresponding circular sec- 
tions of these solids, viz., KL a section of the cylinder, KN 
a section of the sphere, and KM a section of the cone. 

Now FB being equal to FI or IG, and KL parallel to 
FB, then, by similar triangles, IK^KM (B. III., th. 4), and 
IKN is a right-angled triangle ; hence IN 2 is equal to IK 2 
+KN 2 (B.I., th. 34). But KL is equal to the radius IG 
or IN, and KM=IK ; therefore KL 2 is equal to KM 2 +KN 2 , 
or the square of the longest radius of the said circular sec- 
tions is equal to the sum of the squares of the two others. 
Now circles are to each other as the squares of their diam- 
eters or of their radii ; therefore the circle described by KL 
is equal to both the circles described by KM and KN, or 
the section of the cylinder is equal to both the correspond- 
ing sections of the sphere and cone. And as this is always 
the case in every parallel position of KL, it follows that the 
cylinder EB, which is composed of all the former sections, 
is equal to the hemisphere EFG and cone IAB, which are 
composed of all the latter sections. 

But the cone IAB is a third part of the cylinder EB 
(Prop. 10, Corol. 2) ; consequently, the hemisphere EFG is 
equal to the remaining two thirds, or the whole sphere 
EFGH is equal to two thirds of the whole cylinder ABCD. 

Corol. 1. A cone, hemisphere, and cylinder of the same 
base and altitude are to each other as the numbers 1, 2, 3. 

Corol. 2. All spheres are to each other as the cubes of 
their diameters, all these being like parts of their circum- 
scribing cylinders. 



SOLID GEOMETRY. 121 

Corol. 3. From the foregoing demonstration, it appears 
that the spherical zone or frustum EGNP is equal to the 
difference between the cylinder EGLO and the cone IMQ, 
all of the same common height IK. And that the spher- 
ical segment PFN is equal to the difference between the 
cylinder ABLO and the conic frustum AQMB, all of the 
same common altitude FK. 

Scholium. — Bv the scholium to Prop. 10, we have 
cone AIB : cone QIM : : IF 3 : IK 3 : : FH 3 : (FH-2FK) 3 ; 
.-. cone AIB : frustum ABMQ : : FH 3 : FH 3 - (FH - 2FK) 3 
: : FH 3 : 6FH 2 FK- 12FH . FK 2 + 8FK 3 ; 
but cone AIB — one third of the cylinder ABGE ; hence 
cylinder AG : frustum ABMQ : : 3FH 3 : 6FH 2 . FK- 12FH . 

FK 2 -j-8EK 3 . 
Now cylinder AL : cylinder AG : : FK : FI ; 
.-. cylinder AL : frustum ABMQ : : 6FH 2 : 6FH 2 -12FH.FK 

+ 8FK 2 ; 
.-. cylinder AL : segment PFN : : 6FH 2 : 12FH . FK- 8FK 2 ; 

dividendo : : |fH 2 : FK(3FH-2FK). 

But cylinder AL= circular base whose diameter is AB or 
FH multiplied by the height FK ; hence cylinder AL= cir- 
cle EFGHxFK. 

.-.segment PFN=| . circle J;f GH (3FH-2FK)FK 2 . 
F 



SPHERICAL GEOMETRY. 



BOOK VIII. 

DEFINITIONS. 

1. A sphere is a solid terminated by a curve surface, 
and is such that all the points of the surface are equally 
distant from an interior point, which is called the centre of 
the sphere. 

We may conceive a sphere to be generated by the revo- 
lution of a semicircle about its diameter ; for the surface 
described by the motion of the curve will have all its points 
equally distant from the centre. 

2. The radius of a sphere is a straight line drawn from 
the centre to any point on the surface. 

The diameter or axis of a sphere is a straight line drawn 
through the centre, and terminated both ways by the sur- 
face. 

It appears from Definition 1 that all the radii of the same 
sphere are equal, and that all the diameters are equal, and 
each double of the radius. 

3. It will be demonstrated (Prop. 1) that every section 
of a sphere made by a plane is a circle ; this being as- 
sumed, 

A great circle of a sphere is the section made by a plane 
passing through the centre of the sphere. 

A small circle of a sphere is the section made by a plane 
which does not pass through the centre of the sphere. 

4. The pole of a circle of a sphere is the point on the sur- 
face of the sphere equally distant from all the points in the 
circumference of that circle. 

It will be seen (Prop. 2) that all circles, whether great or 
small, have two poles. 



124 SPHERICAL GEOMETRY. 

5. A spherical triangle is the portion of the surface of a 
sphere included by the arcs of three great circles. 
. 6. These arcs are called the sides of the triangle, and 
each is supposed to be less than half of the circumfer- 
ence. 

7. The angles of a spherical triangle are the angles con- 
tained between the planes in which the sides lie. 

8. A plane is said to be a tangent to a sphere when it 
contains only one point in common with the surface of the 
sphere. 

PROPOSITION I. 

Every section of a sphere made hy a plane is a circle 

Let AZBX be a sphere whose cen- 
tre is O. /£ 

Let XPZ be a section made by the x 
plane XZ. 

From O draw OC perpendicular to 
the plane XZ. 

In XPZ take any points P 15 P 2 , P 3 , 




Join CP 1? CP 2 , CP 3 ; also 

OP a , OP 2 , OP 3 ; 

Then, since OC is perpendicular to the plane HZ, it will 
be perpendicular to all straight lines passing through its 
foot in that plane. (Geometry of Planes.) 

Hence the angles OCP l5 OCP 2 , OCP 3 , are right 

angles ; 

/.OP^CP^+OC 2 ; 
OP 2 2 =CP 2 2 +OC 2 ; 
OP 3 2 =CP 3 2 +OC 2 , 
But, since P 1? P 2 , P 3 , are all points upon the sur- 
face of the sphere, .-. by Def. 1, OP 1 = OP 2 =OP 3 = 

.•.CP 1 = CP 2 =:CP 3 

Hence XPZ is a circle whose centre is C, artd every other 
section of a sphere made by a plane may, in like manner, be 
proved to be a circle. 

Corol. 1. If the plane pass through the centre of the 
sphere, then OC = 0, and the radius of the circle will be 
equal to the radius of the sphere. 



SPHERICAL GEOMETRY. 125 

Corol. 2. Hence all great circles are equal to one another, 
since the radius of each is equal to the radius of the sphere. 

Corol. 3. Hence, also, two great circles always bisect each 
other, for their common intersection passing through the 
centre is a diameter. 

Corol. 4r. The centre of a small circle and that of the 
sphere are in a straight line, which is perpendicular to the 
plane of the small circle,. 

'Corol. 5. We can always draw one, and only one, great 
circle through any two points on the surface of a sphere, for 
the two given points and the centre of the sphere give three 
points, which determine the position of a plane. 

If, however, the two given points are the extremities of 
a diameter, then these two points and the centre of the 
sphere are in the same straight line, and an infinite num- 
ber of great circles may be drawn through the two points. 

Distances on the surface of a sphere are measured by the 
arcs of great circles. The reason for this is that the short- 
est line which can be drawn upon the surface of a sphere 
between any two points is the arc of a great circle joining 
them. 

PROPOSITION II. 

If a diameter be drawn perpendicular to the plane of a great 
circle, the extremities of the diameter will be the poles of thai 
circle, and of all the small circles whose planes are parallel 
to it. 

Let APB be a great circle of the 
sphere whose centre is O. 

Draw ZN. a diameter perpendicular 
to the plane of circle APB. 

Then Z and N, the extremities of 
this diameter, are the poles of the 
great circle APB, and all the small 
circles, such as apb, whose planes are 
parallel to that of APB. 

Take any points P l5 P 2 , in the circumference of 

APB. 

Through each of these points respectively, and the points 
Z and N, describe great circles, ZP a N, ZP 2 N. 




126 SPHERICAL GEOMETRY. 

Join OPj, OP 2 , 

Then, since ZO is perpendicular to the plane of APB, it 

is perpendicular to all the straight lines OP 15 OP 2J 

drawn through its foot in that plane. 

Hence all the angles ZOP^ ZOP 2 , . . are right an- 
gles, and .-. the arcs ZP 1? ZP 2 , are quadrants. 

Thus it appears that the points Z and N are equally dis- 
tant from all the points in the circumference of APB, and 
are .-. the poles of that great circle. 

Again, since ZO is perpendicular to the plane APB, it is 
also perpendicular to the plane apb 9 which is parallel to the 
former. 

Hence the oblique lines Zp 19 Zp 2 , drawn top ly p^ 

in the circumference of apb, will be equal to each other. 
(Geometry of Planes.) 

.*. The chords Zp 1? Z^> 2 , being equal, the arcs Z/? 1? 

Z^> 2 , which they subtend, will also be equal. 

.*. The point Z is the pole of the circle apb ; and, for the 
same reason, the point N is also a pole. 

CoroL 1. Every arc PjZ drawn from a point in the cir- 
cumference of a great circle to its pole is a quadrant, and 
this arc PjZ makes a right angle with the arc AP : B. For, 
the straight line ZO being perpendicular to the plane APB, 
every plane which passes through this straight line will be 
perpendicular to the plane APB (Geometry of Planes) ; 
hence the angle between these planes is a right angle, or, by 
{Def. 7), the angle of the arcs APj and Zj is a right angle. 

Corol. 2. In order to find the pole of a given arc AP 15 of 
a great circle, take PjZ equal to a quadrant, and perpen- 
dicular to AP 15 the point Z will be a pole of the ar'c'APj ; 
or from the points A and T 1 draw two arcs AZ and PjZ 
perpendicular to APj, the point Z in which they meet is a 
pole of APj. 

Corol. 3. Reciprocally, if the distance of the point Z from 
each of the points A and Pj is equal to a quadrant, then the 
point Z is the pole of AP 1? and each of the angles ZAP 1? 
ZPjA, is a right angle. 

For, let O be the centre of the sphere, draw the radii OA, 
OP 1; OZ; 

Then, since the angles AOZ, PjOZ, are right angles, the 
straight line OZ is perpendicular to the straight lines OA, 



SPHERICAL GEOMETRY. 127 

OPj, and is ,\ perpendicular to their plane ; hence, by Prop., 
the point Z is the pole of AP 1? and/, the angles ZAP 2 , ZPjA, 
are right angles. 

Corol 4. Great circles, such as ZA, ZP 1? whose planes 
are at right angles to the plane of another great circle, as 
APB, are called its secondaries; and it appears from the 
foregoing corollaries that, 

1. The planes of all secondaries pass through the axis, 
and their circumferences through the poles of their primary ; 
and that the poles of any great circle may always be de- 
termined by the intersection of any two of its secondaries. 

2. The arcs of all secondaries intercepted between the 
primary and its poles are=90°. 

3. A secondary bisects all circles parallel to its primary. 
Corol. 5. Let the radius of the sphere =11, radius of small 

circle =r. Distance of two circles, or Oo=d. 
Join Op!, arc F 1 p 1 =d>. 

r =R cos. (p. 
6 =R sin. <p. 
Corol 6. Two secondaries intercept similar arcs of circles 
parallel to their primary, and these arcs are to each other 
as the cosines of the arcs of the secondaries between the par- 
allels and the ^primary. 

For the arcs of the parallels subtend at their respective 
centres angles equal to the inclinations of the planes of the 
secondaries, and these arcs will therefore be similar. 

Also, if r lt r 2 , be the radii of two small parallels, the rest 
of notation as before, 

circumference p x ^> 2 _whole circumference of 1st 
circumference q l q 2 whole circumference of 2d ' 



r» 



R cos. <f> 
R cos. (// 
cos. cf> 
cos. d/ 



128 



SPHERICAL GEOMETRY. 



PROPOSITION HI. 

Every plane perpendicular to a radius at its extremity is a 
tangent to the sphere in that point. 

Let ZXY be a plane perpendicular to the 
radius OZ. 

Then ZXY touches the sphere in Z. 

Take any point P in the plane ; join.ZP ; 
OP; 

Then, since OZP is a right-angled triangle, 
.*. the side OP is > side OZ. 

Hence the point P is without the sphere; 
and, in like manner, it may be shown that ev- 
ery point in XYZ, except Z, is without the sphere. 

Therefore the plane XYZ is a tangent to the sphere. 




PROPOSITION IV. 

The angles formed by two arcs of great circles is equal to the 
angle contained by the tangents drawn to these arcs at their 
point of intersection, and is measured by the arc described from 
their point of intersection or pole, intercepted by the arcs con- 
taining the angle. 

Let ZPN, ZQN, arcs of great cir- 
cles, intersect in Z. 

Draw ZT, ZT', tangents to the 
arcs at the point Z. 

With Z as pole, describe the arc 
PQ. 

Take O the centre of the sphere, 
and join OP, OQ. 

Then the spherical angle PZQ is 
equal to the angle TZT', and is measured by the arc PQ. 

For the tangent ZT, drawn in the plane ZPN, is perpen- 
dicular to radius OZ. 

And the tangent ZT 7 drawn in the plane ZQN is perpen- 
dicular to radius OZ. 

Hence the angle TZT' is equal to the angle contained by 
these two planes; that is, to the spherical angle PZQ. 
(Geometry of Planes.) 

Again, since the arcs ZP, ZQ, are each of them equal to 
a quadrant, 




SPHERICAL GEOMETRY. 



129 



.-. Each of the angles ZOP, ZOQ, is a right angle ; 

.*. The angle QOP is the angle contained by the planes 
ZPX, ZQX, and is = TZT'. 

.-. The arc PQ, which measures the angle POQ, meas- 
ures the angle between the planes ; that is, the spherical 
angle PZQ. 

Corol 1. The angle under two great circles is measured 
by the distance between their poles. For the axis of the 
great circles drawn through their poles being perpendicular 
to the planes of the circles, the angles under these axes will 
be equal to the angle between the circles ; but the angle un- 
der the axes is obviously measured by the arc which joins 
their extremities; that is, by the distance between their 
poles. 

Corol. 2. The angle under two great circles is measured 
by the arc of a common secondary intercepted between them. 

For, since the secondary passes through the poles of 
both, taking away from the equal quadrants of the second- 
ary between each circle and its pole, the common arc inter- 
cepted between one circle and the pole of the other, the re- 
mainders are the intercept of the common secondary be- 
tween the two circles and the distance between their poles, 
and these are therefore equal. But the latter is, by the last 
corollary, the measure of the angle. 

Corol. 3. Vertical spherical an- 
gles, such as QPW, QPS, are equal, 
for each of them is the angle formed 
by the planes QPS, WPE. w 

Also, when two arcs cut each oth- \ 
er, the two adjacent angles QPTS r , 
QPR, when taken together, are al- 
ways equal to two right angles. 




PROPOSITION v. 

If from the angular points of a spherical triangle considered 
as poles three arcs be described forming another triangle, then, 
reciprocally, the angular points of this last triangle will be the 
poles of the sides opposite to them in the first. 

Let ABC be a spherical triangle. 
F2 




130 SPHERICAL GEOMETRY. 

From the points A, B, C, considered as 
poles, describe the arcs B / C / , A'C 7 , A'B', 
forming the spherical triangle A'B'C. 

Then A" will be the pole of BC, B' of 
AC, and C of AB. -°i 

For, since B is the pole of A / C / , the dis- A 
tance from B to A' is a quadrant. 

And, since C is the pole of A 7 B', the distance from C to 
A' is a quadrant. 

Thus it appears that the point A' is distant by a quad- 
rant from the points B and C. 

.♦. A' is. the pole of the arc BC. 

Similarly, it may be shown that B / is the pole of AC, and 
C the pole of AB. 

PROPOSITION VI. 

The same things being given as in the last proposition, each 
angle in either of the triangles will be measured by the supple- 
ment of the side opposite to it in the other triangle. 

Produce the sides of the first triangle to 
D, E, F, G, H, K 

Then, since A is the pole of B / C / , the 
angle A is measured by the arc EK. 

For the same reason, the angles B and 
C are measured by the arcs DH and FG 
respectively. 

Because B' is the pole of FK, the arc B'K is a quadrant. 

Because C' is the pole of DE, the arc CE is a quadrant. 
/.P/K+C'E^ISO ; or, 
B'C'+EK^ISO ; 

/.EK=180°-B / C / . 

Similarly, DH=180°- A'C ; 

FG=180°-A / B / . 

But the arcs EK, DH, FG are the measures of the an- 
gles A,B, C respectively ; .-. 180°-B / C / , 180°- A'C, 180° 
— A'B', or the supplements of B / C / , A / C / , and A'B' are 
the measures of these angles. 

Again, since A' is the pole of HG, the angle A' is meas- 
ured by GH. 

For the same reason, the angles B', C', are measured by 
the arcs FK and DE respectively. 




SPHERICAL GEOMETRY. 131 

Because B is the pole of A / C, the arc BH is a quadrant. 
Because C is the pole of A / B / , the arc CG is a quadrant. 
.-.BH+CG=:180 o ; or, 
GH+BC=180°; 

.•.GH=180°-BC. 
Similarly, FK=180°-AC ; 

DE = 180°-AB. 
And GH, FK, DE are the measures of the angles A 7 , B', 
C respectively. 

These triangles ABC, A / B / C / , are, from their properties, 
usually called polar triangles, or supplemental triangles. 

PROPOSITION vn. 

In any spherical triangle, any one side is less than the sum 
of the two others. 

Let ABC be a spherical triangle, O 
the centre of the sphere. Draw the ra- 
dii OA, OB, OC. 

Then the three plane angles AOB, 
AOC, BOC form a solid angle at the 
point O, and these three angles are meas- 
ured by the arcs AB, AC, BC 

But each of the plane angles which °" 
form the solid angle is less than the sum of the two others. 

Hence each of the arcs AB, AC, BC, which measure 
these angles, is less than the sum of the two others. 

PROPOSITION VIII. 

The sum of the three sides of a spherical triangle is less than 
the circumference of a great circle. 

Let ABC be any spherical trian- 
gle. 

Produce the sides AB, AC, to meet 
inD. 

Then, since two great circles always 
bisect each other (Prop. 1, Cor.), the 
arcs ABD, ACD, are semicircles. 

Now, in the triangle BCD, 
BC<BD+DC, by Prop. 7 ; 





132 SPHERICAL GEOMETRY. 

.•.AB+AC+BC<AB+BD+AC+DC<ABD-f-ACD, 

< circumference of great circle. 

PROPOSITION IX. 

The sum of all the angles in any spherical triangle is less 
than six right angles, and greater than two right angles. 

For, in the first place, every angle of a spherical triangle 
is less than two right angles ; therefore the sum of all the 
three is less than six right angles. Again, the measure of 
each angle in the spherical triangle is equal to the semicir- 
cumference minus the corresponding side of the polar trian- 
gle (theor. 6) ; hence the sum of all three is measured by- 
three semicircumferences minus the sum of all the sides of 
the polar triangle. Now (Prop. 8) this latter sum is less 
than a circumference ; therefore, taking it away from the 
three semicircumferences, the remainder will be greater than 
one semicircumference, which is the measure of two right 
angles. Hence the sum of all the angles in a spherical tri- 
angle is greater than two right angles. 

Cor oh 1. The sum of the three angles of a spherical tri- 
angle is not constant, like that of the three angles of a plane 
triangle, but it varies between two right angles and six right 
angles, without ever arriving at either of these limits. Two 
angles, therefore, being given, do not serve to determine the third. 

Corol. 2. A spherical triangle may have two or even three 
angles right, two or three obtuse. When it has three right 
angles, it is called a trirectangular triangle ; the surface of a 
sphere, therefore, contains eight trirectangular triangles. 



ELEMENTS OF PLANE TRIGONOMETRY. 



CHAPTER I. 

1. Trigonometry was originally considered to be the 
doctrine of triangles, but in its present improved state it 
has a much more extensive signification. 

2. In estimating angular measures, the right angle is 
supposed to be the primary one, and to be divided into 90 
equal parts, called degrees, marked (°) ; each degree is di- 
vided into 60 equal parts, called minutes, marked (') ; each 
minute is divided into 60 equal parts, called seconds, marked 
(") ; and so on to ( /7 ') thirds, &c. In this division, one de- 
gree is considered as the angular unit. 

3. Modern French writers, instead of using this sexagesi- 
mal division, use the centesimal. 

To reduce French grades into English degrees, or vice 
versa, we use the following formula : 

Let E represent the No. of English degrees, and 
F " " French grades : then 
E F^ ,5__F 

90~100' ° r "9" _ 10 ; 
9 10 

••■ E =fo F > OTF =<? E > 

E = F— [ LF,orF=E+lE. 
Ex. 1. Find the number of degrees and minutes in 46g., 
56' 36". Here F=46.5636 

-±jT = 4.65636 
F— A F=41.90724 
60 



54.43440 
60 



26.064 
Eo = 41o 54' 26"=F— ^ F 



134 ELEMENTS OP PLANE TRIGONOMETRY. 

Ex. 2, Find the number of grades, minutes, &c, in 22° 
27 / 39". 60|39 

60J27.65 

E=22.4608 + 
£E= 2.4956 + 



E+iE=F— 24g. 95' 64". 
4. The circumference of a circle is 3.14159+ times its 
diameter ; or, in other words, the ratio of the circumference 
to the diameter is 3.14159 + . Mathematicians represent 
this number by the Greek letter it* 

:. circumference =ttD ; where D is the diameter; or 

z=2irr, where r is the radius of the circle. 
Hence the length of the arc of a quadrant is i-nr ; of a semi- 
circle, or 180°, it is 7rr ; and of 270°, or three quadrants, it 
is % 717*. Now if any arc, a, subtend an angle of A°, then, 
since ^ixr subtends 90°, and angles are proportional to the 
arcs which subtend them, 

A°: 90 :: a: \%r; 

tt r 
From this expression, any one of the quantities may be 
found when the others are given. 

180° 180° , 72q , 78 

Ex. 3. The earth being supposed a sphere, of which the 
diameter is 7980 miles, find the length of an arc of 1°. 

Ao— 57.29578 x-, or 

lo=57.29578 x 



3990' 



3990 
••-=57Sl= 69 - 6mil - 

Ex. 4. Find the diameter of a globe when an arc of 25° 
of the meridian measures 4 feet. Arts. 18.3346. 



* 77 is also taken to represent 180°, and %ir for 90°. 
Note. — When we take 5 
cular measure of the angle. 



Note. — When we take 57°. 29578 as the unit, - is called the cir- 

r 



ELEMENTS OF PLANE TRIGONOMETRY. 135 

Ex. 5. Find the number of degrees in a circular arc 30 
feet in length, whose radius is 25 feet. 

Am. 68° 45' 17" 44 7// . 

Ex. 6. Find the number of degrees in an angle of which 

the arc, divided by the radius, is .7854, the value of tt being 

3.1416. 

180Q a 180° „„. 180° Ae 

A°= x _ = ^_^x.7854=:-— =45°. 

tt r 3.1416 4 

Ex. 7. The interior angles of a rectilineal figure are in 
arithmetical progression ; the least angle is 120°, and the 
common difference 5° ; required the number of sides. (Ge- 
om.,BookI., Prop. 19.) The sum of all the angles = 2 right 
angles taken (tt — 2) times ; that is (n — 2)n, or 180°n— 360. 
But 120°+ 125°+ 130°+, &c, to n terms=zsum of all the 

angles ; also (Alg., art. 128). A 240+(ra— 1)5 l | = sum of 
all the angles. *• 

Equating these, we. have 

J240 + (tt-l)5| | = 180tt-360 ; 

(240+5tt-5)?i=360tt— 720 ; 
240tt+5tt 2 — 5ttz=360tt-720 ; 
.*. wz=16 or 9. 
The last is the congruent value of n, since no angle can 
be so great as 180° ; .\ the figure has 9 sides. 

Ex. 8. One regular polygon has two sides more than an- 
other, and each of its angles exceeds each angle of the other 
polygon by 15° ; required the number of sides in each. 

Let A = number of degrees in each angle of the 1st polygon, 
A-15= " " " 2d " 

wzzznumber of sides in the 1st, 
n— 2— " " 2d; 

then?iA=(?i— 2)tt (1) 

(»— 2)(A-15) = (w-4)tt .... (2) 

A ns. An octagon or a hexagon. 
Ex. 9. The angles in one regular polygon are twice as 
many as in another polygon ; and an angle of the former is 
to an angle of the latter as 3 to 2 ; to find the number of 
sides. 



136 ELEMENTS OP PLANE TRIGONOMETRY. 

Let A— the number of degrees in the angle of the 1st pol. 
B— « « « 2d " 

rc^ number of sides of the 1st ; 
2n= " " " 2d; 

.:nA=(n-2)n (1) 

2nB = (2?i-2)n (2) 

A:B::2:3 (3) 

Ans. n—4, and 2n=z8. 




(Art. 1.) Definition of the most important Trigonometrical 
Terms. 

1. The complement of an angle is what it wants of being 
90°. Thus, if A be any angle, the complement of A is 
(90° -A). 

2. The supplement of an angle is what it wants of being 
equal to 180°, or to two right angles. Therefore, if A be 
any angle, its supplement will be (180°— A). q 

3. In the annexed diagram, the ratio of PM 
to the radius of the circle is called the sine of 
the angle PAB, and AM to the radius is the 

PM . . AM 

cosine ; or, — — =sin. A, and — — =cos. A. 
AB AB 

4. The ratio of BT to the radius of the circle is called the 

BT 
AB 

5. The ratio of AT to the radius of the circle is called 

AT 

the secant of the angle PAB; or, — — — sec. A. 

AB 

6. The ratio of BM to the radius of the circle is called 

the versed sine of the angle PAB ; or, — — =vers. sin. A. 

AB 

7. The sine of the complement of any angle is called the 
cosine of that angle ; or, 

sin. (90°— A) = cos. A. 
cos. (90°— A) = sin. A. 

8. The tangent of the complement of any given angle is 
called the cotangent of that angle ; or, 

tan. (90°— A)=cot.A. 
cot. (90°— A) = tan. A. 



tangent of the angle PAB; or, -^z^tan. A. 



ELEMENTS OF PLANE TRIGONOMETRY. 



137 



9. The secant of the complement of any angle is called 
cosecant of that angle ; or, 

sec. (90°— A)=cosec. A. 
cosec. (90°— A)— sec. A. 

10. The versed sine of the complement of any angle is 
called the co versed sine of that angle ; or, 

vers. sin. (90°— A) == covers, sin. A. 
covers, sin. (90°— A) = vers. sin. A. 

11. The foregoing definitions relate to an angle less than 
a right angle ; but they are equally applicable to any angle, 
whatever may be its magnitude. For, let 

P'AB be greater than a right angle and qv , 
less than two right angles ; then 

p/M ' • pap 
— — =sm. P AB; 



cos.P'AB; 



AB 

AT 

AB 
B / T / 

— — =tan.P / AB; 
AB 




AT 7 
AB 
B'M 
AB 



^sec.P'AB; 
=:ver. sin. P'AB. 



12. Again: let the angle be situate in the third quadrant 
— that is, greater than two and less than three right angles 
— then we shall have 

P"M" 

:sin.P"AB; 

^cos.P^AB; 
^tan.P^AB; 



AB 
AT" 
AB 

~AB~ 




=sec.P"AB 



:vers. sin. P /r AB. 



13. Finally, let the angle be placed in the fourth quad- 
rant; then 



138 ELEMENTS OF PLANE TRIGONOMETRY. 

p/// M /// 





_ vers. sin. P'"AB; 
AB 

14. The sine of an arc is the perpendicular let fall from 
one extremity of the arc on the diameter passing through 
the other extremity. Thus, MP is the sine of the arc BP ; 
in like manner, AM is the cosine ; BM is the versine ; BT, 
the tangent', AT, the secant, &c, of the arc BP. Hence, 
when the radius is unity, the sine, &c, of the arc become the 
sine, &c, of the angle. 

(Art. 2.) General Relations between Trigonometrical Quantities. 

1. Since AMP is a right-angled triangle, and AP the hy- 

pothenuse, PM 2 +AM 2 =AB 2 . 

Divide by AB 2 , and we have 

PM 2 AM 2 

AB~ 2+ AB 2 

sin . 2 A+cos. 2 A=:i .... (1) 

.-. sin. A== -y/1 —cos. 2 A ; or cos. A= ^/(l— sin. 2 A). 

The triangles PMA, TBA, are similar ; therefore 

PMTB PM PM AM sin. A 

AM~AB ; AM~AB^"AB~c^sTA ; 

sin. A 
hence, r r=tan. A .... (2) 

cos. A 

In equation (2), instead of A, substitute (90°— A) ; 

cos - A 

or, - -—cot. A (3) 

sin. A ' 

Multiplying equations (2) and (3) together, we have 

tan. A. cot. A=l (4) 



-f-T^^ 1 ; that is, 



ELEMENTS OF PLANE TRIGONOMETRY. 139 

AT AP , AM 
A ^ am: AB = AM =1 ^AP ; ° r ' 

sec. A= 



cos. A 



sec. A . cos. A= 1 ; .*. sec. A= r . . (5) 

cos. A 7 



But cosec. A=zsec. (90°— A). \ cosec.A= 



1 



cos. (90°— A)" 
or, cosec. A . sin.A=:l (6) 



sm. A 

AB 2 +BT 2 =zAT 2 . Divide by AB 2 , then 
, BT 2 AT 2 , . 
1 + A^ = A^ ;thatlS ' 
l+tan. 2 A=sec. 2 A (7) 

By squaring equation (3), and adding unity to each mem- 
ber, we have 

cos. 2 A „ . . ' „ . sin. 2 A4-cos. 2 A 

1 +iE?A = 1 + COt - A '-- 1+COt - A = sink ; 

but, sin. 2 A-f cos. 2 A— 1/. l + cot. 2 A=r- — ^-r; hence 

sm. 2 A 

l+cot. 2 A=zcosec. 2 A (8) 

MB AB-AM , AM , 

AB=— AB- = 1 -AB ; henCe 
versin. A=l— cos. A .... (9) 
coversin. A = versin. (90°— A) = 1— cos. (90°— A) ; 
/. coversin. A = l— sin. A . (10) 

11. Trigonometrical quantities are affected by the signs 
plus and minus. 

When the angle A=0, sin. Ar=0. While A varies from 
0° to 90°, sin. A increases from to 1 ; when A=90°, sin. 
A=l. 

While A increases from 90° to 180°, sin. A decreases from 
1 to ; when A=180°, sin. A = 0. 

While A increases from 180° to 270°, sin. A varies from 
0to— 1; when A— 270°, sin. A = — 1. 

While A increases from 270° to 360°, sin. A varies from 
1 to ; when A— 360°, sin. A=0 ; and so on for successive 
revolutions. 



140 



ELEMENTS OP PLANE TRIGONOMETRY. 



The cosine varies in a similar manner, being plus in the 
first and fourth quadrants, and minus in the second and 
third. The variation in the magnitude of the sines and 
cosines being known, the other trigonometrical quantities 
may be easily determined. 

(Art. 3.) Relations existing between the trigonometrical func- 
tions of Angles less than 90°, and those that are greater 
than 90°. 

Draw AP, making any angle A ; let fall 
PM perpendicular on AB ; draw AP 7 , 
making with AC the angle CAP'^PAB 
= A ; and from P' let fall P'M 7 perpendic- 
ular on AB 7 ; then the angle P'AB = 90° 
-f A. The two triangles PAM, P'AM', 
have the side PA of the one equal to P'A 
of the other ; also the angles at M and M' right angles, and 
the angle APM of the one equal to P'AM' of the other ; 
therefore the two triangles are equal in every respect, and 
PM=AM' 5 and AM— P'M'. Divide this last equation by 
AB, and we have 

AM P'M' 

or cos. PAB= sin. P'AB; that is, 




Again : 



AB - AB 

cos.A=sin.(90°+A) . . (11) 

cos. P'AB^sin.PAB: 



AM' PM 



or 



AB AB 

.-. cos. (90°+A)=— sin. A . . . 
But, instead of AP' making an angle 
with AC, let it make that angle with AB ; 
then the angle P'AB is equal to 180° — A. 
The two triangles PAB, P'AB, are mani- 
festly equal in all respects, and PM^P'M', 
AMz=AM'. Divide the first of these equa- 
tions by AB, we have 

PM P'M' . . . /1QAO .. 

; or sin. A=sm. (180°— A) 




(13) 



AB~ AB 

Hence the sine of an angle is equal to the sine of its supplement. 
Again : dividing the other equation by AB, we have 
AM_AM' 
AB~ AB 



■^ = ^± . or cos. A= -cos. (180°-A) (14) 



ELEMENTS OF PLANE TRIGONOMETRY. 141 

That is, the cosine of an angle and the cosine of its supplement 
are equal in absolute magnitude, but have opposite signs. 

If we draw AP', making with AB' an angle equal to A 
in the third quadrant, we shall find, in like manner, 

sin.(180°+A)=-sin.A . . . (15) 

cos.(180°+A)=:Cos.(180 o — A) . (16) 

= — cos. A . . (17) 

If we draw AP', making with AC/ an angle C'APzzzA, 

then sin. (270°— A)= -cos. A . . (18) 

cos. (270°— A)-— sin. A . . (19) 

In like manner we may prove for angles in the fourth 
quadrant. These relations between the sines and cosines 
being established, the corresponding relations between the 
other trigonometrical functions may be deduced immedi- 
ately from the preceding equations. Thus, 

/^n ax sin. (90° + A) cos. A . /0rtv 

ta„(90°+A) = - os / (9()O + A ; = __ = _ C o t .A (20) 

_ „ A sin. (180°— A) sin. A' A „„ 

tan. (ISO"- A)=- ol: i lW; — j=-^^= -tan. A (21) 

And so for all the rest. 

Although the preceding investigations have been confined 
to angles not greater than 360°, still the same principles 
may be applied without any difficulty to angles of any mag- 
nitude whatever. 

Having the sines given, we can calculate the table of co- 
sines, tangents, cotangents, secants, and cosecants by the fore- 
going equations. 

(Art. 4.) To find the sine, cosine, &c, of 30°, 45°, and 60°. 

Let the angle be 60°. In the equilateral £ 

triangle ACB, let fall the perpendicular CD, 
which bisects both the angle ACB and the 
base AB. (Geom., B. I., Prop. 3.) ^ 

AD 
AC 

1AC 



Now, xr^sin. ACD=sin. 30° ; but AD=-|AC ; 



/.sin. 30°: 



AC 



cos. 30°= VI -sin. 2 30° = ^/(l ~i)= i/i=W% 5 but 



142 ELEMENTS OF PLANE TRIGONOMETRY. 

cos. 30°=sin. 60° /. sin. 60°=^B 5 
cos. 60°=:sin. 30° .-. cos. 60° =% ; 

..tan.dO -^300-1^3-^3' 
, o^o cos. 30° -i-v/3 

,■ SeC '^ U -cos.30°-iV3-V3' 

cosec. 30°= . ^o = T = 2 ' 
sm. 30° -J 

In like manner, find tan., cot., sec, and cosec. of 60°. 
When the angle is 45°, 

sin. 45o— cos. (90°— 45°)=:Cos. 45°. 

Now, since cos. 2 45°-f sin. 2 45° = l, and cos. 45°=:sin. 45°, 
we have 2 cos. 2 45°= 1/. cos. 2 45°= \; 

or cos. 45° = - v /-i=iV 2 i 

.♦.sin. 45 =-iy 2 » 
u sin. 45o ^y2 

hence tan. 45°= — -=rf-^=l; 

cos. 45o Jy 2 

cos. 45° , 

COt. 450=-—- — = 1; 

sm. 45° 

sec. 45°= —= T - r —<x/2; 

cos. 45° -iy^2 v 

cosec. 45°= . AKO = V 2 - , 
sin. 45° v 

Ex. 10. If tan. #=§, find the sm. and versin, and cos. #. 

sin. x sin. a; . 

tan. xz= /. =z£ ; 

cos. x cos. x 

3 sin. #=4 cos. x; 

=4-^/(1 — sin. 2 a); 

9 sin. 2 #=16 — 16 sin. 2 a;; 

25 sin. 2 x=i 16 .*. sin. #= : | ; 

cos. #=f , vers. x=% . 

jEfe. 11. Given 6 sin. 2 as =5 cos. a, to find sin., cos., and 
tan. x. Ans. sin.x=^^/5; 

cos. a:— § or — |; 
tan. x=^^/5. 



ELEMENTS OF PLANE TRIGONOMETRY. 143 

Ex. 12. Given sin. x cos. x—^^/'d, to find sin. x and cos. x. 

Ans. sin. x—\-\fiS or \\ 
cos. x—\ or \^§. 

4 
i£c. 13. Given tan. a?+cot. x=— — , to find tan. x and 

V 3 1 

cot. x. Ans. tan. #:= V3 or — j- ; 

1 V° 
cot. a; =-^5 or -y/3. 

i£c. 14. Given 3 sin. x-\-5-\/3 . cos. a:=9, to find sin. #. 

Ans. sin. aj=-J- or -£-. 

Ex. 15. Given sin. #-f-sin. ?/= — - — 

sin. a; sin. 2/ =^^3, to find sin. x and 
sin. y. Ans. sin. x=^; 

sin.?/ =-^3. 
Ex. 16. Prove that sec. 2 a?cosec. 2 a;=sec. 2 a:-{-cosec. 2 a;. 
Ex. 17. Given tan. 2 a:-|-4 sin. 2 #=6, to find sin. x. 

Ans. sin. x—^^/d. 
Ex. 18. Given sin. #(sin. x — cos. x)=^ 5 , to find sin. x. 

Ans. sin. x==%. 
Ex. 19. Given 6 tan. a;-}- 12 cot. #=:5^/3 . sec. a?, to find 
tan x. Ans. tan. x= -y/3. 

Ex. 20. Given tan. a?-f-cot. x=2, to find tan x. 

Ans. tan. x—1. 
Ex. 21. Given tan. 3 #+7 tan. #=22, to find tan. a?. 
Multiply each member of the equation by tan. x, and we 
have tan. 4 £-{-7 tan. 2 ic=22 tan. x; 

add 4 tan. 2 a? to both members, and 

tan. 4 a;-}- 11 tan. 2 £=4 tan. 2 a; -|- 22 tan. a:; 
comp. D? an( l extract the root, and reduce, we have 
tan. x=2. 

Ex. 22. Given tan. 2 x4-\=~ U't cot. a?, to find tan. x. 

8 tan. a; b ' 

tan. 3 #+-§• tan. x=: 1 -|~|. = |. ; 

multiply by tan. x, and we have 

tan. 4 x-f^ tan. 2 #:=:-§ tan. x. 

If we now add tan. 2 x to both members, we can complete 

the square and extract the root. Ans. tan. x=?l. 

Ex. 23. Given 4 sec. 2 £=195 cot. a;— 117, to find tan. x. 

Ans. tan. #=•§. 



144 ELEMENTS OF PLANE TRIGONOMETRY. 

Ex. 24. Given sin. 4 x— 2 sin. 2 ^—1 = 2 sin. x— cos. 2 a, to 
fifid sin. x. Ans. sin. x-=. — 1. 




CHAPTER II. 

PROPOSITION I. THEOREM I. 

(Art. 5.) 1. In any right-angled triangle, the ratio which the 
side opposite to one of the acute angles bears to the hypothe- 
nuse is the sine of that angle ; 2. the ratio which the side adja- 
cent to one of the acute angles bears to the hypothenuse is the 
cosine of that angle ; 3. and the ratio ivhich the side opposite 
to one of the acute angles bears to the side adjacent to that 
angle is the tangent of that angle. 
Let ABC be any plane triangle right-angled at B ; 

, CB . . AB . ^ 

then — —sin. A; or— — =sin. C . (1) C ' 

AL AC c 

AB ■ BC n 

— =cos. A; or ^=cos. C . (2) 

BC A AB „ /0 **" 

— =tan.A; or — =tan. C . (3) 

Proof. — From A as a centre, with radius AC, describe 
an arc of a circle ; produce AB to meet the circumference in 
B' ; from B' draw B / C / , a tangent to the arc at B' ; pro- 
duce AC to meet B / C / in C' ; then, from definitions (3) 
and (4), 

BC . . AB A A B ' C ' + A 

^7=sin. A, ^g7=cos. A, and —, =tan. A. 

-r>p AB 

But AB 7 - AC, whence — — =sin. A ; — ==cos. A. 
AC AC 

And the triangles ABC, AB / C / , being similar, 

B'C BC BC 

AC=AB-'-AB= tan - A - 

proposition n. theorem h. 

In any plane triangle, the ratio of any two of the sides is equal 
to the ratio of the sines of the angles opposite to them. 
Let ABC be a plane triangle ; it is required to prove that 




ELEMENTS OP PLANE TRIGONOMETRY. 145 

CB sin. A CB sin.A 
CA~sin. B ; BA~sin. C ; 
CA sin. B 
BA~~sin. C* 
Draw CD perpendicular to AB ; then, by theorem 1, 

DP DC 

— - = sin. A ; ^^ = sin. B. Divide the first of these equations 
CA OB 

DC DC sin. A CB sin. A 

by the second, we have y^r-r- 7^5— - — =5, or -— - —- — — . 

J CA CB sin. B CA sm. B 

In like manner, by letting fall perpendiculars from B and 

A upon the sides AC, CB, we can prove that 

CB_sin.A CA__sin.B 

BA~sin7C ; BA~sm7C' 

In treating of plane triangles, we make use of the capital 
letters A, B, C to designate the three angles, and the corre- 
sponding small letters a, h, c to represent the sides opposite to 
them. According to this notation, this last theorem will be 
a sin. A a sin. A b sin. B 
5~sin. B ' c ~ sin. C ' c — sin. C' 



(Art. 6.) Investigation of General Formulas, 

PROBLEM I. 

Given the sines and cosines of two angles, to find the sine of 
their sum. 

Let ABC be a plane triangle (see the last figure) ; then, 
by equation 2, theorem 1, 

BD=BC cos. B, AD=zAC cos. A ; but 
AB=BD+ AD =BC cos. B+ AC cos. A. 

Divide each term of this equation by AB, we obtain 

BC AC 

1=-^=: cos. B+ -r-=r cos. A ; but by theorem 2, 
AB AB J 

C sin. A j AC sin. B 
•~i = - — ft? an d Tn =- — 7i- Substitute these in 
AB sin. C AB sm. C 

,, , , sin. A cos. B sin. B cos. A 

the above, we have 1= 7 — 1 : — - — ; 

sm. C sm. C 

.*. sin. C=sin. A cos. B+sin. B cos. A ; 

a 



146 ELEMENTS OF PLAKE TRIGONOMETRY. 

but, since ABC is a plane triangle, A+B+C=180 o .\ C 
= 180° — (A+B); and sin. C=sin. jl80° — (A+B)} = sin. 
(A+B), because 180°— (A+B)isthe supplement of (A+B). 
.*. sin. (A+B)=sin. A cos. B+sin. B cos. A . . (1) 
Equation (1), from its great importance, is called the fun- 
damental formula of Plane Trigonometry, and nearly the 
whole science may be derived from it. 

problem n. 

To find the sine of the difference of two angles, the sines and 
cosines of the two angles being given. 

By equation (1), sin. (A+B)=sin. A cos. B-f-sin. B cos. 
A; for A substitute (180°— A), the above will become 
sin. {(180°-A)+B}=sin. {180o— (A-B)} =sin.(180° 

—A) cos. B+sin. B cos. (180°— A). 
But sin. {180°— (A— B)} -sin. (A-B), 

and sin. (180°— A) = sin. A ; cos. (180°— A) = —cos. A. 
Substitute these in the above expression, it becomes 

sin. (A— B)=sin. A cos. B— sin. B cos. A . . (2) 

problem in. 

Given the sines and cosines of two angles, to find the cosine of 
their sum. 

Formula (1), sin. (A+B) = sin. A cos. B+sin. B cos. A; 
for A substitute (90°+ A), we have 

sin. {(90°+A)+B;=sin.(90o+A) cos.B+sin.B 
cos. (90o+ A). 
But sin. {9Qo+(A+B)}=cos.(A+B),and 

sin. (90°+A)=r:cos. A, and cos. (90°+ A)= —sin. A. 
Substituting, therefore, these values in the above expression, 
it becomes 

cos. (A+B)=cos. A cos. B— sin. A sin. B . (3) 

PROBLEM IV. 

To find the cosine of the difference of two angles. 

Formula (1), sin. (A+B)— sin. A cos. B+sin. B cos, A ; 
for A substitute (90°— A), it will become 

sin. {(90°-A)+Bi=sin. {90°-(A-B)j =sin. (90° 
-A) cos. B+sin. B cos. (90°- A). 



ELEMENTS OF PLANE TRIGONOMETRY. 147 

But sin. {90°-(A-B)j = cos. (A-B) ; 

sin. (90°— A)— cos. A ; 

cos. (90°— A)=:sin. A. 
Substitute, therefore, these values in the above expression, it 
becomes 

cos. (A— B)— cos. A cos. B+sin. A sin. B . . (4) 

PROBLEM V. 

To find the tangent of the sum of two angles. 

We have by Chap. I., equation 2, 

, . . _ sin. (A + B) _ sin. A cos. B -f sin. B cos. A 

' ~~ cos. (A -f- B) ~" cos. A cos. B — sin. A sin. B' 

Divide both the numerator and denominator of this fraction 

by cos. A cos. B, we shall have 

sin. A cos. B sin. B cos. A 

/* ™ cos. A cos. B cos. A cos. B 

tan. (A+B)= : — . 

- - sin. A sin. B 

cos. A cos. B" 

Simplifying, 

.. _. tan. A+tan. B 
tau.(A+B )= - tan _+ tanB .... (5) 

PROBLEM VI. 

Given the tangents of two angles, to find the tangent of their 
difference. 

(Chapter I., equation 2) : 

' . ^ K sin. (A— B) sin. A cos. B— sin. B cos. A 

tan. (A— B)=: ~ — =^= - — : — z — '- — — . 

cos. (A— B cos. A cos. B+sin. A sin. B 

Dividing both the numerator and denominator by cos. A 

cos. B, it becomes 

sin. A cos. B sin. B cos. A 

- . cos. A cos. B cos. A cos. B 

tan. (A — B) = : — 7 —-. — = 

sin. A sin. B 

cos. A cos.B* 

Simplifying, we have 

/K tan. A— tan. B , 

tan. (A— B)=:- - A - .... (6) 

v ' 1 4- tan. A tan. B v ' 



148 ELEMENTS OF PLANE TRIGONOMETRY. 

The sum of the equations (1) and (2) is 

sin. (A -f- B) + sin. (A — B) = 2 sin. A cos. B . (7) 
Their difference is 

sin. (A+B)— sin. (A+B)=2 sin. B cos. A . (8) 
The sum of (3) and (4) is 

cos. (A+B)*-j-cos. (A— B)=2 cos. A cos. B . (9) 
Their difference is 

cos. (A+B)— cos. (A+B)m2 sin. A sin. B . (10) 
If, in equations (7), (8), (9), and (10), we make 
A+Bz^p, and A— B=g, that is, 

A=±(p+q) B=±(p — q), we shall have 
sin.p-\-sm. q =2' sin. ±(p -\-q) cos. ±(p—q) . (11) 
sin.p— sin. q—2 sin. \{p— q)cos. ^(p-\-q) • (12) 
cos. p-f- cos. 2=i 2 cos. ^(p~\-q) COS. ^(p — q) . (13) 
cos. q— cos.p — 2 sin. ±(p-{-q)sm. i(p— q) . (14) 
Divide (11) by (12), we obtain 

sin. p -f- sin. q sin. \{p -f- q) cos. ±(p — q) 

sin. p— sin. q~cos. \(p-\-q) sin. ^(p—q)' 

Eeducing the second member, it becomes 

sin. »4-sin. a ,". . , , . tan. \(p-\-q) ,„ ^ v 

. F ^ . * =tim.±(p+q)cot.l;(p-q )=- ^~~. (15) 

sm.p—sm.q 2Vjr ' a/ 2XM H tan.iQ)— #) v ' 

Dividing (14) by (13), we have 

cos. q — cos. p 

^— — =tan. Up~\-q) tan. i(p—q) ; 

cos. q-\- cos. p : JXI 1} 2Vi v 

__ tan. -j(j?4-g) _ tan. -^ j>— g) 
~cot.iQ9-2)~cot. i(i?+g) 
(15) may also be written 

sin. p -}- sin. g tan. ^(j^-j-g) cot. \(p — q) 
sin. p— sin. q~ tan. J(^— g) — cot. i(/>+g) 
If, in (13) and (14), we make g=0, we shall have 
l-f-cos.p=2 cos. 2 -^, andl — cos.^>=:2 sin. 2 ^ . . {a) 
sin.^+si n.g = sin. i(i?+g) ^ tan> i( , v ( „ 
cos.^-fcos. # cos.J(^» + ^) 2\F-T\L) ' V / 

sin.^+sin.g cos. |(j>— gj 

=-: T7 r = COt. Mp — g) . (c) 

cos. q— cos.p sm. ^(p—q) ^ v x/ v ' 

sin.p— sin. # sin. i(p— g) , , 

~ - = |r— -~-i^n.\{p—q) . (d) 

cos.p -f cos. q cos. %(p—q) ■ v 2/ v ' 

sin. p— sin. q cos.h{p-\-q) , n , 

cos. q— cos. |? sin. £(/>+#) v v/ 



(16) 
(17) 



ELEMENTS OP PLANE TRIGONOMETRY. 149 

cos. j>+cos. g _ cos. j(p+q) cos. j(p — g )_ cot. j(p+q) . 
cos. q^—cos.p~ sin. ^(^+$) sin. J^— q)~ tan. i(^>_ g) ^ ' 
&m.pr\-sm.q__sm.i(p-{-q) cos.%(2)—q)_cos.%(p—q) 

sin. (iJ+g) _ sin. J(i>4-?) cos. i(p+q)~ cos. £(.?>+?) ^' 
sin. ^— sin. g_sin. ^ (_£>—#) cos. ^( £>+#)_ sin. \(p—q) 

sin. (i?+?) ~ sin. i(l>+2) cos - UP+<l)~ sin - KP+?) 

PROBLEM YTI. 

(Apt. 7.) To determine the sine of tivice a given angle. 
Formula 1 : 

sin. (A-j-B)=:sin. A cos. B-f-sin. B cos. A. 
Let Bz=A, then the equation becomes 

sin. 2A=sin. A cos. A-f sin. A cos. A 

=2 sin. A cos. A (18) 

Cor. In (18), just obtained, let ^A be substituted for A; 
then sin. Az=2 sin. JA cos. £A ..'.'. . (19) 

PROBLEM YTJff. 

To determine the cosine of twice a given angle. 
By formula (3) : 

cos. (A+B)=cos. A cos. B— sin. A sin. B. 
Let B=A, the above becomes 

cos.2A=cos. 2 A~-sin. 2 A . . (20) 
But sin. 2 A = 1 — cos. 2 A. Substitute this for sin. 2 A, then 

cos.2A=2cos. 2 A— 1 . . . (21) 
Again: since cos. 2 A =1— sin. 2 A, we have 

cos. 2A=l-2 sin. 2 A . . . (22) 



PROBLEM IX. 

By formula (5), of the present chapter : 

.. • tan. A-}- tan. B 

tan. (A+B): 



1— tan. A tan.B 
Let Bin A, the above becomes 

*-*HS& • • • • < 23 > 

In like manner deduce the following : 
problem x. 

cot . 2 A = ^=- 1 . . . . (24) 

2 cot. A v ' 



150 ELEMENTS OF PLANE TRIGONOMETRY. 



sec.2A= , . (25) 



PROBLEM XL 




sec. 2 A cosec. 


2 A 


~~ "cosec. 2 A— sec 


. 2 A 


PROBLEM XII. 




sec. 2 A cosec. 2 


A") 



cosec. 2A=- j ^— r-f . (26) 

2 sec. A cosec. A r v : 

==j sec. A cosec. A J . (27) 

PROBLEM XIII. 

To determine the sine of half a given angle. 

By Formula (22) : 

cos.2A=l— 2 sin. 2 A. 

For A substitute £A, and we have 

cos. 2 . (£A) = 1— 2 sin. 2 iA; 

or cos. A = 1 — 2 sin. 2 ^A ; 

2 sin. 2 i A =1— cos. A; 

. 1— cos. A 
sin. 2 £A= ; 



. -i * /l— cos. A __. v 

sm-iA^y^ 2 ' ' (28) 



PROBLEM XIV. 

To determine the cosine of half a given angle. 
By Formula (21): 

cos. 2Az=2 cos. 2 A-l. 
For A substitute \ A, we have 

cos. A =2 cos. 2 JA— 1 ; 
or 2 cos. 2 JA= 1+cos. A; 



, , /1 + cos. A ynn , 

cos. JA^-t^ . . . (29) 



PROBLEM XV. 

To determine the tangent of half a given angle. 

Divide equation (28) by (29)j 

sin. -AA /l— cos. A 



iA VI 



cos. \ A V 1 + cos. A' 

, a /l— cos. A /ft . x 

or tan.JA=\/— r . . (30) 

A V 1+cos. A ' 



ELEMENTS OF PLANE TRIGONOMETRY. 151 

Multiply both numerator and denominator of the fraction 

by yl— cos. A, we have 

1— cos. A /01 . 

tan.£A= — : — r— .... (31) 

1 sin. A. 

Multiply both numerator and denominator of (30) by 

1+cos. A, we have sin. A 

tan. £A= 1+cos. A 

In like manner deduce 

/1 + cos. A 1+cos. A sin. A 

cot. \K—\- r == — : — r — =-z r- (33) 

1 V 1— cos. A sin. A 1— cos. A 

. . /2sec. A t 

sec. iA=\/ rT1 (34) 

* V sec. A+l ' 



cosec. JA: 



/ 2 sec. A 
ViSTA=i ' (35) 



PROBLEM XVI. 



To determine the sine of (ra+l)A, in terms of wA, (n— 1)A, 
arad A. 
Equations (1) and (2) may be thus written: 

sin. (B+ A)=sin. B cos. A+sin. A cos. B ; 
sin. (B— A)=sin. B cos. A— sin. A cos. B. 
Add these two equations, and we have 

sin. (B+A)+sin. (B— A) = 2 sin. B cos. A. 
Subtract sin. (B— A) from each member, then 

sin. (B+A)=2 sin.B cos. A— sin. (B— A). 
Let B=wA, then 

sin. (ra+l)A=2 sin. nA cos. A— sin. (n— 1)A . (36) 
In the above formula, let n=l ; then n-\-l=z2, n— 1=0. 
Therefore sin. 2A=2 sin. A cos. A— sin. ; 

—2 sin. A cos. A, the same as equat. (24). 
Letrc=2; then n-\- 1=3, and n — 1 = 1. 
Therefore sin. 3A=2 sin. 2 A cos. A— sin. A ; 

=2x2 sin. A cos. A cos. A— sin. A ; 
=4 sin. A cos. 2 A— sin. A ; 
=4 sin. A (1— sin. 2 A)— sin. A; 
=4 sin. A — 4 sin. 3 A— sin. A ; 
=3 sin. A— 4 sin. 3 A . . . (37) 
Let n=3 ; then rc+l=4, and n— 1=2. 
Therefore sin. 4A=2 sin. 3 A cos. A— sin. 2 A ; 
but, by the last equation, sin. 3 A =3 sin. A— 4 sin. 3 A. 



152 ELEMENTS OF PLANE TRIGONOMETRY. 

Hence 2 sin. 3A=2(3 sin. A— 4 sin. 3 A), and 

sin. 4A=(6 sin. A— 8 sin. 3 A) cos. A— 2 sin. A cos. A; 

= {(6—8 sin. 2 A) cos. A— 2 cos. A} sin. A; 

— j(6_(8— 8 cos. 2 A)cos. A— 2 cos. AJsin. A; 

= {—2 cos. A-f-8 cos. 3 A— 2 cos. A} sin. A; 

= {8 cos. 3 A— 4 cos. A} sin. A ... (38) 
In like manner we may continue the process for sin. 5 A, 
sin. 6A, &c. 

problem xvn. 
To determine the cosine of (n+l)A in terms ofnA, (n—l)A, 
and A. 
By equations (3), (4) : 

cos. (B-j-A) = cos.B cos. A— sin. B sin. A; 
cos. (B — A)=cos. B cos. A-f-sin. B sin. A. 
Adding these two equations, we have 

cos. (B+A) + cos. (B— A) =2 cos. B cos. A. 
Subtract cos. (B — A) from both members, then 

cos. (B-|-A)=2 cos.B cos. A— cos. (B— A). 
Let B=wA, we have 

cos. (n-\- 1)A=2 cos. nA cos. A— cos. (n— 1)A . (39) 
Put w=l, then ?z-|-l = 2, n— 1=0. 
Therefore cos. 2 A =2 cos. 3 A— cos. ; 

=2 cos. 2 A— 1, the same as (27). 
Put w=2, then n-\-l=S, n— 1 = 1, and 

cos. 3A=2 cos. 2 A cos. A— cos. A; 

=2 (2 cos. 2 A— 1) cos. A— cos. A; 
=4 cos. 3 A— 2 cos. A— cos. A; 
=4 cos. 3 A— 3 cos. A . . . . (40) 
Put w=3, then rc+l=4, n— 1=2. 
cos. 4A=2 cos. 3 A cos. A— cos. 2 A ; 

=2(4 cos. 3 A— 3 cos. A) cos. A— (2 cos. 2 A— 1); 
=8 cos. 4 A— 6 cos. 2 A— 2 cos. 2 A+l; 

=8 cos. 4 A- 8 cos. 2 A+l (41) 

In like manner we may find cos. 5 A, cos. 6 A, &c. 
(Art. 8.) By adding and subtracting (1) and (2), (3) and 
(4), we will have the following useful formulas : 
sin. (A-f B)+sin. (A— B)=2 sin. A cos.B 
sin. (A-fB)— sin. (A— B)=2 sin. B cos. A 
cos.(A+B)+cos.(A-B)=2 cos. A cos.B 
cos. (A+B)— cos. (A— B)= — 2 sin. A sin. B 
Multiply (1) and (2) together, we have 



ELEMENTS OF PLANE TRIGONOMETRY. 153 

sin. (A+B) sin.(A-B)=sin. 2 A cos. 2 B-sin. 2 B cos. 2 A. 
But 

cos. 2 B = l-sin. 2 B.\ sm. 2 A cos. 2 B = sin. 2 A-sin. 2 A sin.'B; 
and 

cos. 2 A=l-sin. 2 A.-. sin. 2 B cos. 2 A = sin. 2 B— sin 2 A sin. 2 B. 
Substitute these values in the above equations, then 
sin. (A-f B) sin. (A— B) -sin. 2 A— sin. 2 A sin. 2 B— 
(sin. 2 B— sin. 2 A sin. 2 B); 
hence sin. (A+B) sin. (A— B)=sin. 2 A-sin. 2 B . (43) 

Multiply (3) and (4), we shall have 
cos. (A+B) cos. (A-B)=cos. 2 A cos. 2 B-sin. 2 A sin. 2 B. 
But 

cos. 2 B=l— sin. 2 B.\ cos. 2 A cos. 2 B = cos. 2 A— cos. 2 A sin. 2 B; 
and 

sin. 2 A=l— cos. 2 A.\ sin. 2 A sin. 2 B=sin. 2 A-cos. 2 Asin. 2 B. 

Substitute these values in the above equation, and we have 

cos.(A+B) cos. (A-B)=cos. 2 A-cos. 2 A sin. 2 B- (sin. 2 A 

—cos. 2 A sin. 2 B); 
hence cos. (A+B) cos. (A— B)=cos. 2 A— sin. 2 A; 
or cos. (A+B) : cos. A+sin. A : : cos. A— sin. A : cos. 
(A-B) (44) 

THEOREM III. 

(Art. 9.) In any plane triangle, the sum of any two sides is to 
their difference as the tangent of half the sum of the 
opposite to them is to the tangent of half their difference. 

(Theorem 2.) T — ~. * ; or a: b — sin. A: sin. B. 
v b sin. B 

By the theory of proportions, we have 

a-\-b: a— &=sin. A+sin. B : sin. A— sin. B. 

a-\-b sin. A+sin.B 



Whence 



a — b sin. A — sin. B* 



/-o x- -i>t\ sin.A+sin.B tan.J(A+B) 

(Equation 17) : - — A . „ =- ?; A -J ; or, 

v * ' sm.A— sm.B tan.^(A— B) ' 

g+5_ tan. j(A+B) 
a—b~tsm. ^(A— B) 

T ,., a+c tan. *(A+C) 

In like manner, — L -==- :; A ' V . . . . (45) 

a— c tan. £(A— C) 

6+c_tan. i(B+C) 

£^c = tan. i(B^C") 

G2 



154 



ELEMENTS OP PLANE TRIGONOMETRY. 



THEOREM IV. 

In any plane triangle, the greatest side is to the sum of the other 
two sides as their difference is to the difference of the segments 
made by a perpendicular let fall from the angle opposite the 
greater side. 

Demonstration. Let ABC be a plane tri- 
angle, of which the side AB is the greatest. 
With the centre C and radius =BC, describe 
the circle FEBG. Draw CD perpendicular 
to AB. Produce AC to meet the circle at 
G; drawCE. 

Then, because AB and AG- are secants, it will be 
As AB : AG : : AF : AE (Geom., B. IT., Prop. 21). But 
AG=AC+CG=AC+BC, and AF=AC-CF=AC-~ 
BC, and AE= AD— DE= AD— DB. Substitute these val- 
ues of AG, AE, and AF, in the above proposition, we have 
AB:AC+BC::AC-BC:AD-DB . (46) 




\^X J 



THEOREM V. 

In any plane triangle, the cosine of either of the angles is equal 
to the sum of the squares of its adjacent sides minus the 
square of the side opposite divided by twice the rectangle of 
the adjacent sides. 

1°. Let ABC be a triangle, A, B, C the 
angles, and a, b, c the sides respectively. 
Let A, the proposed angle, be acute. From 
C draw CD perpendicular to AB ; then 

BC 2 = AC 2 + AB 2 -2AB x AD (Geom., B. I., Prop. 37) ; 
or a l =¥-\-c l — 2cxAD. 

But, because CDA is a right-angled triangle, 
AD= AC cos. A—b cos. A ; 
.-.a 2 =& 2 +c 2 -2fccos.A; 




\ cos. A= 



5 2 +e 2 



2bc 
2°. Let A, the proposed angle, be obtuse : 
from C draw CD perpendicular to AB pro- 
duced; then D j± 



ELEMENTS OF PLANE TRIGONOMETRY. 155 

BC 2 =AC 2 + AB 2 +2BA x AD ; 

or a 2 =5 2 +c 2 +2cxAD. 

But, since CD A is a right-angled triangle, AD=AC cos. 
CAD.-. AD = AC x —cos. CAB (because CAB is the sup- 
plement of CAD). 

.*. AD = —b cos. A; hence a 2 =5 2 -}-c 2 — 25c cos. A; 

or cos. A= — 

2bc 

The two results are identical. Therefore, whether A be 
acute or obtuse, we shall have 

& 2 +c 2 -a 2 

C0S - A= — 25c— < 47) 

To express the sine of an angle of a plane triangle in terms of 
the sides of the triangle. 
Let A be the angle ; then, by (47), 

cos. A= — . 

25c 

Add unity to each member of the equation, 

i , a i ■ & 2 +c 2 -a 2 

l+cos.A=l+— ^—; 

5 2 +25c+c 2 -a\ 
~~ Wc ; 

(b+c+a ){b+c-a) 

= 2bc ' * (48) 

Again : subtracting each member of (equation 47) from 
unity, we have 

1-COS. A=l — ; 

2bc 

_ 2bc-V-c*+a\ 
~ 2bc '' 

_ a 2 -(5 2 -25c+c 2 ) ^ . 

~ 2bc ; 

_ (a + o- C )(a +C -b) 

~ 2bc ' ' ' ^ 

Multiply equations (48), (49), we shall have 

1 -cos. 2 A=sin. 2 A= ( a + & +c)(ft+c-a)(a+ft-c)(a+c-a) 

45 V 



156 ELEMENTS OF PLANE TPJGONOMETKY. 

.\ sin. A=-j- V(a+o+c)(b+e-a){a-{-b-c)(a-\-c- b) (50) 

The above equation may be expressed somewhat differ- 
ently, thus : 

Let 5— \(a-\-b+c) ; then 2s=a-\-b-\-c. Now subtract 2a 
from both members, we have 

2s— 2a—b-\ r c— a; 
b-\-c—a 



or s—a= 



In like manner, s— b= 



2 ' 

a+c— b 



i— c=- 



2 
a-\-b—c 



2 

Substitute 2s, 2(s— a •• ), &c, in the expression for sin. 2 A, 
it becomes 

-;„ ^ a __ 16fl(a-fl)(s--ft)(8-c) 

2 

therefore sin. A=— y^s— a)(s— £)(s— c) . . . (51) 

To express the sine, cosine, and tangent of half an angle of a 
plane triangle in terms of the sides of the triangle. 

(a-\-b + c)(b + c—a) 



(Equation 49) : 1+cos. A= 



2bc 
4ts(s—a) 



2bc 
but 1 -j- cos. A = 2 cos. 2 JA 

hence 2 cos. 8 JA=^=^ ; 

1 2bc 



^co S .jA=y / 53 



be 
In like manner, 



4^V(^); • • • ( 53 > 



Again (25) : 1 —cos. A =2 sin. 2 ^A ; 



ELEMENTS OP PLANE TRIGONOMETRY. 

4(s-b)(s—c) 



157 



.•.2sin. 2 *A: 



hence 

In like manner, 



sin. \A. 



^fcfc)) 



By dividing (52), (53), (54), by (49), (50) 



(55) 

(56) 

(57) 
(51), we obtain 
(58) 

(59) 

(60) 



CHAPTER in. 

(Art. 10.) Construction of Trigonometrical Talks. 

It is manifest, from Definitions (1), (2), (3), &c, that the 
various trigonometrical quantities, the sine, cosine, tangent, 
&c, are abstract numbers, representing the comparative 
lengths of certain lines. In Chapter II. the numerical val- 
ues of some of these quantities have been found : it is the 
purpose of this to show how the numbers corresponding to 
angles of every degree of magnitude may be obtained by the 
application of simple principles. 

The first operation to be performed is 

1°. To compute the numerical 

i 
The arc of one minute is 



of the sine ofY. 

of the semi-circumfer- 

10800 

ence, whose radius is unity. The length of this semi-cir- 
cumference has been found to be 3.1415926 -f-. Therefore 

the ^l'=!^|gp = o.qod290888 . . (61) 



158 ELEMENTS OF PLANE TRIGONOMETRY. 

sin. 2'=?^^ =0.000581776 . . (62) 
5400 v ' 

&c.= &c.= &c. 

sin.lo = 3 - 14 l 1 8 5 o 926 = 0.01745 . . . (63) 

2°. Having the sine of 1°, to find the sine of 2°, 3°, 4°, 5°, 6°, 
7°, 8°, 9°, 10°. 
Equation (18) : 

sin. 2A=:2 sin. A cos. A =2 sin. AVi — sin. 2 A. 
Put A=l° ; then 

sin.2°=2 sin. l°Vl-sin. 2 l°=0.034895 . (64) 
Equation (40) : 
sin.(A+B) sin. (A— B)=sin. 2 A-sin. 2 B; 

=:(sin. A -{-sin. B)(sin. A — sin. B); 

or, sin. (A— B) : sin. A— sin. B : : sin. A+sin. B : sin. (A+B). 

That is, the sine of the difference of any two angles is to the 

difference of their sines as the sum of those sines is to the sine 

of their sum. 

Therefore, put A=2o, B=l°; then 
sin. 1° : sin. 2°-sin. 1° :: sin. 2°+sin. 1° : sin. 3°=.052336. 

Put A=3°,B=1°: 
sin. 2° : sin. 3°— sin. lo :: sin.[3°+sin. 1° : sin. 4 o =.069756. 
By assuming A =4°, 5°, &c, B=l°, we obtain the sines 
required. Using the same proportions, knowing the sines 
of 5°, 10°, and 15°, the sines of 20°, 25o, 35°, 55°, 65o, & c ., 
may be found. 
Equation (25) : 

j (~\_ cos A_\ t — - 

sin. 4A=y ^ -— j =0-4 VI -sin. 2 A. 

Substitute ^A for A, and it becomes 

sin. £A, or sin. \tK—^J%— J VI— sin. 2 ^A; 

sin. i 3 A=r yjl~ i Vl -sin. 2 i 3 A ; 

and generally, sin. £» A = w ^ — \ V 1 — sin. 2 \n-\ A. 

Assume A =30°, then £A=15° ; and applying the above 
formula, we have 

sin. 15°=y / J-iVl-sin. 2 30 o . 



ELEMENTS OF PLANE TRIGONOMETRY. 159 

But sin. 30° ==§, and sin. 2 3Qo=l; 

.-. sin. 15°= > yj-iv'l I= i=0.258819. 
In like manner, 

sin. 7° 30'^y^-^-v/l-sin. 2 lP 



=0.1305268, 
&c, &c. 
3°. Again, from equation (1): sin. (A+B)=sin. A cos. 
B-f-sin. B cos. A. 
Let B=l'; then 

sin. (A-fl'^sin. A cos. l'-f-sin. 1/ cos. A ; 
but cos. l'=l and sin. l'r^O.00029. Therefore, 

sin. (A-j-l'^sin. A+sin. Y cos. A ") , fi -. 

=sin. A+. 00029 cos. Aj * K > 
By equation (65) the sine of an angle of any number of 
degrees and minutes may be calculated. 

, (Art. 11.) To find the sine, $c, of 0° and 90°. 

Formula (1) : sin. (A— B)=sin. A cos. B— sin.B cos. A, 
Put B=:A ; we have 

sin. 0°=sin. A cos. A— sin. A cos. A=0 ; 
sin. 0° = cos. 90°=0. 
Formula (4) : cos. (A— B)=cos. A cos. B+sin. A sin. B. 
PutB=A; then 

cos. 0°=cos. A cos. A -J- sin. A sin. A ; 

=cos. 2 A+sin. 2 A=l; 
cos.0°=sin. 90°= 1. 

tan.0°=cot.90° =^! = ?=0; 
cos. 0° 1 

cot.0°=tan.90° 

sec. 0°=cosec. 90°= 



cosec 0°=sec. 90° =-. — ^=-=:oo . 
sm. 0° 



cos. 0° 

~sin. o= 
1 


1 

"0" 
1 


= oo; 

=1; 


~~cos. 0°" 
1 


~1~ 

1 



160 ELEMENTS OF PLANE TRIGONOMETRY. 

To find the sine of 180°. 

Formula (24) : sin. 180°=2 sin. 90° cos. 90°=0 (72) 
(26) cos. 180 o =cos. 2 90 o -sin. 2 90=:-l . (73) 

— «" ES--J— ■■ « 5 > 

(Art. 12.) To find the sine, &c, 0/270°. 

In formula (1), make A=180°, B=90° ; then 
sin. (180°+90°)=sin. 180° cos. 90°+sin. 90° cos. 180°; 

sin. 270° = -1* (78) 

cos. (180°+90°)=cos. 180° cos. 90°-sin. 180° sin. 90°; 
cos. 270°=0 (79) 

t a n.270-"^-i=-ao. . (80) 
cos. 270° 0. ' 

cot . 270 ° = ^§™! = , . . {g 

sm. 270° —1 v 

sec - 270 °=S3T¥70-o= s=« • • (81) 
cosec - 270O =ihTW=-I=- 1 - • ( 82 > 

To find the sine, $c, o/360°. 

In formula (18), make A=180° ; then 

sin. 2 A =2 sin. A cos. A becomes 
sin.360° = 2 sin. 180° cos. 180°=0 (83) 
(26) cos.360°=cos. 2 180 o -sin. 2 180 o = l=cos.0° (84) 

Therefore the sine, &c, of 360° are the same as those 
of0°. 

To find the sine, §c, of an angle greater than 360°. 
By formula (1) : 
sin. (360°+A)=sin. 360° cos. A+sin. A cos. 360° 

=sin.A (85) 



ELEMENTS OF PLANE TRIGONOMETRY. 161 

(Art. 13.) Hence the sine, &c.,, of an angle greater tha» 
360° are equal to the sine, &c., of its excess above 360°. 

To find the sine, fyc, of a negative angle- 

In formula (2), make A=0 ; then 

sin. (— B)=sin. cos. B— sin. B cos. 
= — sin. B. v 
&c, &c. 

QUESTIONS FOR EXERCISE. 

1. If be any angle, prove that tan. 2 0— sin. 2 0=:tan. 2 

sin. 2 0. But 1 —sin. 2 0=cos. 2 (from Eq. 1, Chap. I.). 

Multiply each member of this equation by sin. 2 6, and it 

becomes 

sin. 2 0-sin. 2 sin. 2 0=sin. 2 cos. 2 ; 

or, by transposition, > 

sin. 2 0— sin. 2 cos. 2 0=sin. 2 sin. 2 0. 

Divide by cos. 2 0, we have 

sin. 2 . sin. 2 . „ . 
-sm. 2 0:= r-^ sm. 2 0. 



cos. 2 cos. 2 

sin. 2 

cos. s 
. tan. 2 0— sin. 2 (9=tan. 2 sin. 2 0. 



-d * sm. 2 

But — -=tan. 2 

cos. 



2. Prove the trigonometrical formulas : 

sin. 40=sin. £0(2 cos. 0+1) ; 
cos. |0=cos. 10(2 cos. 0-1) ; 

tan.f9= sin - e + sin -l|g. 

2 cos. 0+ cos. 20 

1°. sin. §0=sin. (0+ i0)=sin. cos. 10+sin. 10 cos. ; 
r=2 sin. £0 cos. 2 ^0+ sin. 10 cos. 0; 
=sin.£0(2 cos. 2 £0+ cos. 0). 
But 2 cos. 2 J0=1 + cos. 0; 

/. sin. f 0=sin. £0(2 cos. 0+1). 
2°. cos.-§0=cos.(0+£0)=cos. cos. 10— sin. sin. £0; 
=cos.0cos.£0— 2 cos.£0sin 9 2 £f« 
=cos. £0(cos. 0—2 sin. 2 £0). 
But 2 sin. 2 10=1 -cos. 0. 

/. cos. -|0=cos. £0(2 cos. 0—1). 



162 ELEMENTS OF PLANE TRIGONOMETRY. 

30 tan 3g_ sin -f^ sin -M 2 cos.0+l) 
* ¥ cos.-§0 cos. ±6(2 cos. 6— 1) 
_ 2 sin. jL0 cos. jd(2 cos. +1) 
~" 2 cos. 2 ^0(2 cos. 0—1) 

sin. (9(2 cos. 0+1) 
~~(l+cos.0)(2cos.0-l) 

sin. 0+2 sin. cos. sin. 0+sin. 20 
" cos. 0+2 cos. 2 0—1 ~~cos. 0+ cos. 20* 

3. Eliminate from the two following equations, and 
find the value of x. 

x cos. 0+a sin. 0—5 .... (1) 
x sin. 6— a cos. 0=c .... (2) 
Square the two equations, and we have 

x* cos. 2 d+2ax sin. cos. 0+a 2 sin. 2 0=5 2 . (3) 
a 2 sin. 2 6—2ax sin. cos. 0+a 2 cos. 2 0=c 2 . (4) 
Add (3) and (4) together, we have 

a 2 (sin. 2 0+cos. 2 0)+a 2 (sin. 2 0+cos. 2 6)=P-\-cK 
But sin. 2 0+cos. 2 0-1. 

.•.a: 2 +a 2 =5 2 +c 2 ; 
whence x == V# 2 + c^—a*. 

4. In a plane triangle, given 

5=a sin. C (1) 

c=a cos. B (2) 

to find the angles. 

Multiply equation (2) by 2c, we have 

2c 2 =2«ccos.B .... (3) 
By (47) : 5 2 =a 2 +c 2 — 2ac cos.B; add these, we have 

&+<*=a* (4) 

Hence the triangle is right-angled, a being the hypothe- 
nuse; then A=90°, and B+C=90°; 

sin. C=cos. B; 
b=a sin. C, c=a cos. B ; 

Hence the triangle is isosceles.-. A =90°, B= 45°= C 
=45°. 

5. If be any angle, prove that 

tan.^-tan.He= 8sin " 9 8 r 19 
* cos. 2 

Solution : 

tan. 2 0— tan. 2 £0= (tan. 0+tan. £0)(tan. 0— tan. |0) ; 



ELEMENTS OF PLANE TRIGONOMETRY. 163 

, • sin. sin. W 

but, tan. 0+ tan. Id— h -\ ~ ; 

1 3 cos. 6 cos. ^0 

sin. cos. ^-0-f-cos. sin. J0 
cos. cos. ^0 

cos. cos. ^0 ' 
sin.|-0 

/ cos. cos. -|-0 ' 

4 sin. ^-0 cos. -J-0 cos. §0 # 

cos. cos. ^-0 
4 sin. ^0 cos. §0 

cos. 

. _ sin. sin. 4-0 

and tan. 0— tan. W= 5 r^ ; 

d cos. cos. ^0 

sin. J0 



cos. cos. ^0 ' 
_2 sin. J0 

— COS. 

Therefore taa.'0-W^ 8 ^'^^ ig. 

* cos. 2 

6. Prove that the sum of the tangents of the three angles 
in every plane triangle is equal to their product. 

Solution. If 0, 0', 0" be the three angles, then + 0' -f 6" 
= 180°; 
or tan. (18O°-0)=tan. (0'+0"). 

Therefore — tan. — - — j , ' ; 

1— tan. tan. 7/ 

hence —tan. 0+tan. tan. 0' tan. 0"=tan. 0'+tan. 0" ; 

or tan. 0+tan. 0'+tan. 6"=tan. tan. 0' tan. 0". 

7. Given a sin. 0+5 cos. 0=c, to find 0. 

. . fl ac±hVc?+b' i -c 1 
Ans. sm. 0= — — . 

a 2 + 6 2 

8. If the tan. of an arc be equal to -y/n, prove that the 
sine of the same arc is=\ / . 

9. Eliminate from the two equations. 



164 ELEMENTS OF PLANE TKIGONOMETKY. 

x cos. 0-|-a sin. 6—h, 
x sin. 0-j-a cos. d—c. 

{x+ay {x—ay 

10. Given tan. 0-f-cot. d—Tc, to find 0. 

.4ws. tan.0=J±i-\/^ 2 _4. 

11. Given k cos. 0=tan. 0, to find 0. 



Ans. cos.0=y -^- 2 ±~ 2 V4F+1, 



12. Prove the formula 

tan. ^04- cot. \Q—2 cosec. 0. 

13. Find the angle </>, from the equation 

1+acos. (04-0) l+acos.0 - . 

: — . , _/. — : — -7 — , and and a being known. 

sin. (04-0 ) sin. & 

^ * il * */ , cos. (0-00 

.Ams. tan. M=cot. 4- a r — « — -• 

/r sm. / 

14. In any plane triangle ABC, if the angle made by a 
line drawn from the angle C to the middle point of the side 
c be denoted by 0, prove that 

2 cot. 0=cot. A— cot. B. 

15. If the angles A, B, C of a plane triangle be as the 

a-\- c 
numbers 2, 3, 4 respectively, prove that 2 cos. ^A=— — . 

(Akt. 14.) The following formulas, which are sometimes called 
the formulas of verification, are very useful. 

Since cos. 2 a 4- sin. 2 a— 1, 

and 2 sin. a cos. a=sin. 2a, 

by adding and subtracting, we have 

cos. 2 «-J-2 sin. a cos. a4-sin. 2 a=l-f-sin. 2a, 
and cos. 2 a — 2 sin. a cos. a+sin. 2 a=l — sin. 2a. 

By extracting the square root of each of these, we have 
cos. a-fsin. a= -^/(1 + sin. 2a) ; 
cos. a— sin. a= -y/(l —sin. 2a). 
When a is less than 45°, sin. a is less than cos. a ; there- 
fore, by adding, 

2 cos. a= -^/(l + sin. 2a) -f- -\/(l— sin. 2a) ; 
.-. cos.a=J-v/(l+sin. 2a) -j- \ -\/(l— sin. 2a). 



ELEMENTS OF PLANE TIUGONOMETKY. 165 

Subtracting, 2 sin. a= -/(l +sin. 2a) — -y/(l —sin. 2a) ; 
.-. sin. a=^-/(l4-sin. 2a)— ^V(l— sin. 2a). 
When a is greater than 45°, then sin. a is less than cos. a, 

and the above will become 

cos.a=%{^(l + sm.2a) — -/(l — sin. 2a)} . . (A) 
sin.a=i{V(l+sm.2a)+y(l-sin.2a)} . . (B) 

1. To find the sine of 18°. 

Since the cosine of an angle is equal to the sine of its 
complement, cos. 54°=sin. 36°; but 

cos. 54°=cos. (36° + 18°)=cos. 36° cos. 18°— sin. 36° 
sin. 18° ; 
/. cos. 36° cos. 18°— sin. 36° sin. 18°= sin. 36° ; 
but cos. 360=1 — 2 sin. 2 18°, 

and sin. 36° =2 sin. 18° cos. 18°. 

.-. cos. 18°(l-2 sin. 2 18°)— 2 sin. 2 18° cos. 18°= 
2 sin. 18° cos. 18°; 
.-. 1-2 sin. 2 18°-2 sin. 2 18° =2 sin. 18°; 
.-. 4 sin. 2 18°+2 sin. 18°=1. /B? , 

., sin .i8°=^-=i 

4 

2. Prove that cos. (60° + a) -f cos. (60°— a) = cos. a. 
1°. cos. (60°+a)=cos. 60° cos. a— sin.' 60° sin. a. 
2°. cos. (60°— a)=cos. 60° cos. a+sin. 60° sin. a. 

;. cos. (60° + a) + cos. (60°— a) — 2 cos. 60° cos. a 
=2 : X:i cos. a=cos. a; 
because cos. 60°=^. (Trig., Art. 4.) 

3. Prove that sin. (30°+a)+sin. (30°— a)=cos. a. 
2 tan. \a 



Prove that sin. a= 



l + tan. 2 £a 



2 C 

5. Prove that 2 cos. (45°+ Ja) cos. (45°— Ja)=cos. a. 

6. Given sin. (x-\-a) + cos. (a?+a) = sin. (x— a) + cos. 
(# — a)j to find sin. x. Ans. sin. »= J-y^2. 

7. Given sin. sc+cos. 2#= J-y/5, to find sin. x. 

a • V5- 1 

-4?2S. Sin. iCrz:- 1 — . 

4 

8. Given 4 sin. a? sin. 3a* =1, to find sin.#. 

An, S in.*=^. 

n -d ,l x 2 sin * «— sin. 2a . 

9. Prove that 7—: = — :r- =tan. 2 *a. 

2 sin, a + sin. 2a 



\ 

166 ELEMENTS OP PLANE TRIGONOMETRY. 

10. IfflH-J+c=90°, prove that 

cot. a -{-cot. 5-j-cot. c=cot. a cot. b cot. c. 

11. If a+b+c=180°, prove that 

tan. a-}- tan. b-\- tan. c=tan. a tan. b tan. c. 

12. Find the tangent of 15°. 

* iKo * /ko qaon tan.45°-tan.30° 

tan. 15°=tan. (45°— 30°)=— — — j- ^r- 

v ' l+tan.45° tan. 30° 

1 "V8 _ V3-1 
+ V3 



CHAPTER IV. 

INVERSE TRIGONOMETRICAL FUNCTIONS. 

(Art. 15.) The quantities, sin. -1 a, cos. -1 a, &c, are called 
inverse functions, and have the following signification : 

(1.) x—^mr 1 a is read, x is equal to an arc or angle whose 
sine is a. 

(2.) £=cos. -1 a is read, x is equal to an arc whose cosine 
is a. 

(3.) a? = tan. -1 a is read, x is equal to an arc whose tan- 
gent is a. 

From (1) : sin. x=a ; (2) cos. x— a ; (3) tan. x=z «. 

Let sin. a=p, and sin. b—q; 

then a = sin. -1 />, and b — sin. -1 <? ; 

and, since sin. a=p, ,\ cos. a— -\/l — p\ cos. 5= -y/l — 2 2 ? 
sin. (a +6)= sin. a cos. 5 -(-cos. a sin. b ; 

=J>V(l-<f)+2"/(l-/)- 
.-. (a+^^sin.- 1 {p^/(l-.f) +q ^/{l-f))., 

but a = sin. -1 j>, and 5= sin. -1 #. 

.-. sin.-^+sin.- 1 g=sin. -1 {p^/(l — 2*)+sV(l — p 2 )}. 

In like manner, 

sin. -1 p — sin. -1 q = sin. -1 { p -y/( 1 — <f) — # -y/( 1 — p 2 ) } . 
By proceeding in a similar manner for the cosine, 

cos. -1 p -f cos. -1 q == cos. -1 { ^ — -/( 1 — p 2 ) -y/( 1 — <f) } ; 

cos. -1 p— cos. -1 q=z cos. -1 {£>£+ V (1— £> 2 ) -j/(l— <f )j. 



ELEMENTS OF PLANE TRIGONOMETRY. 167 

Proceeding in the same manner for the tangents : 

If tan. a=p, and tan b = q, 

then a=tan. -1 £>, and 5=rtan. -1 q ; 

tan. a 4- tan. b p4-q 

tan. ( a +b)=——±- % =±~~- 

v ' 1— tan. a tan. b 1—pq 

V—PV 

But «=tan. -1 £>, and 5=tan. -1 q ; 

/. tan. -1 »+tan. -1 ? =ztan. -1 (i^L). 

\l-pq) 

, . _. tan. a— tan. b p—q 

And since tan. (a— b)=— — - = * , ; 

1-ftan. a tan. # l+i?2 

.*. tan. -1 p— tan. -1 <7=tan. -1 ( f- — - J. 

In like manner, _ 

cot. -1 » ± cot. - q = cot. -1 | — ) . 

\ ?=»=/>/ 

1. If #=sin. -1 (■§), and ?/z=zsin. -1 (|), then #+?/=90 o . 
For sin. »=■§; cos. #=1/(1 — A)=f; 

and sin.#=f; cos. y= y(l_^f)=f ; 

sin. (a; +?/):= sin. a; cos.?/-|-cos. a; sin. y; 

3 V 3 l 4 \s±» 

9 |_ 1 6 1 

2. Prove that 
tan. -1 -^4-tan." 1 i+tan. -1 -i +tan. -1 ^=45°. 

(tan. -1 -^+tan. -1 -J) -{-(tan. -1 -f-ftan. -1 -J) = 1st mem. ; 

3ut tan. -1 +4- tan. -1 i=tan. -1 i Y ,, f- =tan. -1 f , 



3 . 



tan. -1 J+ tan. -1 £= tan. -1 ] 1 5 1 [ =tan. 
r i F 
and tan. -1 i+tan. -1 -I^tan. -1 ] J Yi \ = tan - _1 

.-. tan. - ^+tan. -1 -J-+tan. -1 -l-+tan. -1 -|-=ztan. -1 1'+tan." 1 -^-. 
But 

tan. -1 f +tan. -1 -^-=ztan.- 1 {-l±i-l = tan. -1 1=45^ 

3. Prove that tan. -1 -f+2 tan. -1 J=45°. 

4. Given tan. -1 ( J —tan. -1 ( J — — -, to find x. 



^ ^yfcr 



168 ELEMENTS OP PLANE TRIGONOMETRY. 

5. Prove that 

cc cc 

6. Given sec. -1 a— sec. -1 b= sec. -1 T — see. -1 -, to find on, 

b a 

Ans. x—±^db. 



SOLUTION OF 

BIGHT-ANGLED -PLANE TRIANGLES 

BY MEANS OF LOGARITHMS. 



The solution of right-angled triangles may be embraced 
in four cases : 

1°. When the hypothenuse and one of the acute angles are 
given. 

Chap. II., Theor. I. By multiplying the first member by 
radius to make the equation homogeneous, 
R.BC=AC.sin.A(l), 

/. (a.) As radius is to the sine of the acute angle, so is the hy- 
pothenuse to the side opposite ; and, 

(b.) As radius : the cosine of the acute angle : : the hypothenuse 
to the side adjacent. 

2°. When the hypothenuse and one side are given. 

From the above equation (1) we have, 

(c.) As the hypothenuse : the given tide : : radius : the sine of 
the angle opposite the given side. 

(d.) As radius: the sine of the angle opposite the required side 
: : the hypothenuse : the side required. 

3°. When a side and one of the acide angles are given. 

(Chap. II., Theor. I.) 

BC=AB.tan.A, 
or R.BC^AB.tan.A. 

(e.) As radius: the tangent of the given angle:: the side adja- 
cent : the side opposite the given angle. 

(/.) As sine of the angle opposite the given side : radius : : given 
side : the hypothenuse. 

4°. When the two sides are given. 

Using the same formula as the last, we have, 

(g.) As the base : the perpendicular ; : radius : the tangent of 
the angle at the base. 

H 



170 SOLUTION OF RIGHT-ANGLED PLANE TRIANGLES. 

(h.) As sine the angle at the base: radius:: the perpendicular 
: the hypothenuse. 

EXAMPLES. 

1. In the right-angled triangle ABC are given the hypoth- 
enuse AC=250, and the angle A=73° 44' 23", to solve 
the triangle. Ans. C = 16° 15' 37". 

«z=:240. 

c=70.0003. 

2. Given the angle A = 50° 30', the hypothenuse = 
142.27, to solve the triangle. Ans. C=39° 30'. 

a=109.78. 
c= 90.5. 

3. Given the hypothenuse = 47.467, and one side == 
37.29, to solve the triangle. 

Ans. Other side =29.37. 

A , f88°13'27". 
Angles | 51 o 46 , 33 - 

4. Given the perpendicular —2 22, and the angle opposite 
the base =25° 15', to solve the triangle. 

Ans. 64o 15', 104.7, and 245.45. 

5. Given the base=123, and perpendicular =765, to solve 
the triangle. Ans. 80° 51' 57". 

9° 8' 3". 
774.82. 



SOLUTION OF 

OBLIQUE-ANGLED PLANE TRIANGLES. 



The solution of all oblique-angled plane triangles may be 
included under four cases : 

1°. When one side and two angles are given. 

Since the sum of the three angles of every plane triangle 
is equal to 180°, therefore, if we subtract the sum of any 
two angles of the triangle from 180°, we shall have the third 
angle. 

Then, from Chap. II., Theor. II., we have the following 

RULE. 

As sine ofjlie angle opposite the given side : sine of the angle 
opposite the required side : : the given side : required side. 

2°. When two sides and an angle opposite one of them are 
given. 

From the same theorem as above we derive the following 

RULE. 

As the side opposite the given angle : the other given side : : 
sine of the given angle : the sine of the angle opposite the other 
given side. 

Having found a second angle, the third is obtained as 
above, and the side is found by Case 1°. 

3°. When two sides and their included angle are given. 

First, subtract the included angle from 180° for the sum 
of the other two angles. Then, by Theor. III., we have the 
following 

RULE. 

As the sum of the two given sides : their difference : : tangent 
of half the sum of the opposite angles : tangent half their differ- 
ence. 

Then to half the sum of the two angles add half their dif- 



172 SOLUTION OF OBLIQUE-ANGLED PLANE TRIANGLES. 

ference for the greater angle, and from half their sum take 
half their difference for the less angle. 

The third side is found by Case 1°. 

4°. When the three sides are given. 

The student may use any of the formulas under Theorem 
V., from 50 to 60, for the solution of the problem ; but per- 
haps from Theorem IV. we may have a more convenient 

RULE. 

Consider the longest side the base. Then, 

As the base : the sum of the other two sides : : the difference of 
those sides : the difference of tlie segments made by a perpendicu- 
lar lei fall from the vertical angle. 

Then, to half the base add half the difference of the segments 
for the greater segment, and from half the base take half the 
difference of the segments for the less segment. Now, 

As the greater of the other two given sides of the triangle : the 
greater segment : : radius : the cosine of the less angle ; and, 

As the less of the two given sides : the less segment : : radius 
: the cosine of the greater angle at the base. 

EXAMPLES. 

1. In the triangle ABC are given the side c= 93.37, the 
angle A=30° 20', and the angle C=99° 30', to find the 
other parts. Ans. a— 47.81. 

5 = 72.697. 
B=50° 10'. 

2. In the triangle ABC are given the side c=364, the 
angle A=57° 15', and the angle B=35° 30', to find the 
other parts. Ans. a =306.49. 

5=211.62. 
C=87° 15'. 

3. In the triangle ABC are given the side a= 5 1.234, 
the side 5=42.356, and the angle A=55°, to find the oth- 
er parts. Ans. c=61.992. 

B=42°37'33". 
C=82°22'27". 

4. In the triangle ABC are given the side 5=50.24, the 
side c=43.25, and their included angle A=40° 15', to find 
the other parts. 

Ans. B=81° 24' 25", C=58° 20' 35", and «=32.829. 



SOLUTION OF OBLIQUE-ANGLED PLANE TRIANGLES. 173 

5. Given the side a=133, the side c=176, and their in- 
cluded angle B = 73°, to find the remaining parts. 

Ans. A=42° 50' 57". 
C=64° 9' S". 
6 = 187.022. 

6. In the triangle ABC are given the side a=50.25, the 
side 6=60.5, and the side c=68.4, to find the angles. 

Ans. A=45° 22' 35". 
B=58°58'18". 
C=75°39' 7". 

7. Given «=150, 6=140, and c=130, to find the angles. 

Ans. A=67°22' 48". 
B=59° 29' 23". 
C=53° V 49". 

8. Given a=. 8706, 6 = .0916, and c=.7902, to find the 
angles. Ans. A= 149° 49'. 

B= 3° 1'56". 
C= 27° 9' 4". 

9. Given a=31.23879, 6=49.00117, and A=32° 18', to 
find the other parts. Ans. B= 56° 56' 56". 

C= 90° 45' 4". 

c= 58.456. 
Or, B=123° 3' 4". 

C= 24° 38' 56". 

c=24.381. 



MENSURATION. 



The term mensuration is frequently applied to a system 
of rules and methods by which numerical measures of geo- 
metrical quantities are obtained. It is to this limited view 
of the subject that the present chapter shall be confined. 

In all practical applications of mathematics, it is neces- 
sary to express magnitudes of every kind by numbers. For 
this purpose, a line of some determined length, as one inch, 
one foot, etc. , is assumed as the measuring unit of lines, 
and the number expressing how often this unit is contained 
in any line is the numerical value or measure of that line. 

A surface of some determinate figure and magnitude is 
assumed as the measuring unit of surfaces, and the number 
of units contained in any surface is the numerical measure 
of that surface, and is called its area. It is usual to as- 
sume, as the measuring unit of surfaces, a square, whose 
side is the measuring unit of lines. 

A solid of a determinate figure and magnitude is, in like 
manner, assumed as the measuring unit of solids, and the 
number of units contained in any solid is its solidity or vol- 
ume. The unit of solids is a cube, each of whose edges is 
the measuring unit of lines, and, consequently, each of its 
faces the measuring unit of surfaces. The Tables of Square 
and Cubic Measure may be found in Arithmetic (Art. 126, 
127). 



MENSURATION OF SURFACES. 

PROBLEM I. 

To find the area of a parallelogram, square, or rectangle. 
From (Geom., B. IV., Prop. 3, Cor. 2), we have the fol- 
lowing 



MENSURATION, 175~ 

RULE. 

Multiply the base by the altitude, and the product will be the 
area. 

EXAMPLES. 

1. Find the area of a parallelogram, the base being 15 
chains and the altitude 10.50 chains. 

Ans. 15 acres, 3 roods. 

2. Find the area of a square each side of which is 80 
chains. Ans. 640 acres. 

3. Find the area of a rectangle whose base is 100 feet 
and altitude 75 feet. Ans. 7500 square feet. 

PROBLEM II. 

To find the area of a triangle whose base and altitude are 
given. 

From (Geom., B. IV., Prop. 3, Cor. 3), we have the fol- 
lowing 

RULE. 

Multiply the base and altitude together, then half the product 
will be the area. 

EXAMPLES. 

1. Find the area of a triangle whose base is 5 chains and 
altitude 3.50 chains. Ans. 3 roods, 20 perches. 

2. What is the area of a triangle whose base is 40 rods 
and altitude 8 rods ? Ans. 1 acre. 

problem in. 

To find the area of a triangle ivhen its angles and one side 
are given. 

Solution. By (prob. 2), the area of the 
triangle 

ABC=iCD X AB ... (1) 

/.Areaof ABC=^^5. 

But, by (th. 2, Trig.), c= a * m '^ , and (th. 1) CD=asin.B. 
sin. j\. 

Substitute these in (equation 1), we have 

Area of ABC = — — - - — z ; 

2 sin. A 




176 MENSURATION. 

o a e a nn a2 sin * B sin * C 
or 2, Area oi ABC = : : . 

sin. A 
Or, in logarithms, 

Log, double area=2 log. a-flog. sin. B-f-log. sin. C-f-ar. 
co. log. sin. A — 20. 

From which we derive the following 

RULE. 

To twice the logarithm of the given side add the logarithmic 
sines of the adjacent angles, and the arithmetical complement q 
the logarithmic sine of the opposite angle, and reject 20, andth^ 
sum will be the logarithm of double the area of the triangle. 

examples. 

L Given one side of a triangle equal 24.32 chains, and 
the angles adjacent 63° and 74°, to find the area. 
Solution. 

a=24.32 twice log. 2.771928 

13 = 63° sin. 9.949881 

C = 74o sin. 9.982842 

A =43° ar. co. sin. 0.166217 

Double area=742.793 sq. chs. 2.870868 

.-. Area=37 acres, 22 perches. 
2. Given a side equal 19 chains, and the adjacent angles 
60° and 50°, to find the area. 

Ans. 12 acres, 2 roods, 39 perches. 

PROBLEM IV. 

To find the area of a triangle when two sides and their in- 
cluded angle are given. 

Solution. Area of ABC = \c x CD. 

But, by (th. 1, Plane Trig.), DC = a sin. B ; 
.*. Area of ABC — hac sin. B 
2 Area ABC=oc sin. B . . . (1) ' 
Or, in logarithms, 

log. double area = log. a-\-\og. c-f-log. sin. B . (2) 
Cor. The area of a parellelogram is equal to ac sin. B (3) 
From (1) we perceive tiiat the double area of a triangle is 
equal to the product of the two sides into the natural sine of the 
included angle. And from (2) we obtain the following 



MENSURATION. 177 

RULE. 

Add together the logarithms of the two sides and the loga- 
rithmic sine of the included angle, reject 10, and the remainder 
will be the logarithm of double the area of the triangle. 

EXAMPLES. 

1. Given the side 5 = 15.36 chains, the side c= 11.46 
chains, and the included angle A =47° 3CK, to find the area. 
Here #=15.36 log. 1.186391 

c=11.46 1.059185 

A=47° 30' sin. 9.867631 



Double area=129.78 sq. chs. 2.113207 
.-. Area=6 acres, 1 rood, 38 perches. 
2. Given the side a = 18.23 chains, the side 5=13.84 
chains, and their included angle C = 66° 30', to rind the 
area. Ans. 11 acres, 2 roods, 11 perches. 

PROBLEM V. 

To find the area of a triangle when the three sides are given. 

From (prob. 4), 

Area of ABC=^ac sin. B. 
But, by (equation 51, Plane Trig.), 
2 



sin. B= - Vs (s—a) (s—b)(s—c) ; 
ac v ' y ' 



.\ Area of ABC = \ac sin. B= -\/ s (s—a) (s—b) (s—c). 

Or, in logarithms, 

Log. area ABC =^ {log. s+log. (s— a)+log. (s— 5)-f-log. 
(s-c)}. 

In which s denotes the half sum of the three sides of the 
triangle. From which we derive the following 

RULES. 

From half the sum of the three sides subtract each side sep- 
arately ; then multiply the half sum and the three remainders 
together, and the square root of their continued product will be 
the area of the triangle. Or, 

RULE II. 

Add together the logarithms of the half sum and the three re- 
H 2 



178 MENSURATION. 

mainders and half the sum of these logarithms will be the log- 
arithm of the area. 

EXAMPLES. 

1. Given the three sides of a triangular field equal to 
21.28, 22.56, and 18 chains respectively, to find the area. 

Here s=30.92; s-a=9.64; s-b=8.3Q; s-c=zl2.92; 
,\ by Eule I., 

V30.92 x 9.64 x 8.36 x 12.92 = 179.429 sq. chs. 
Or, by Logarithms, Rule II., 

s=30.92 log. 1.490239 

s-a=9.64 0.984077 

s-b=8.36 0.922206 

s-c=12.92 1.111263 

2 )4.507785 

log. area=2.253892 

,\ Area =179.429 sq. chs.= 

17 acres, 3 roods, 30.8 perches. 

2. Given the three sides of a triangular field equal to 49, 
50.25, and 25.69 chains respectively, to find the area. 

Ans. 61 acres, 2 roods nearly. 

PROBLEM VI. 

To find the area of a trapezoid. 

RULE. 

Multiply half the sum of the two parallel sides by the alti- 
tude of the trapezoid. (Geom., B. I., th. 29.) 

Or, Multiply the altitude by the distance between the middle 
points of its inclined sides. 

EXAMPLES. 

1. What is the area of a trapezoid the parallel sides of 
which are 8 and 12 chains respectively, and the perpendic- 
ular distance between them 5 chains ? 



m= 



50 sq. chs. =5 acres. 



2. What is the area of a trapezoid the parallel sides of 
which are 12.25 and 7.50 chains, and its altitude 15.40 
chains? Ans. 15 acres, 33 perches. 



MENSURATION. 179 

The area of a trapezium or of any irregular polygon may he 
found by drawing diagonals, and thus dividing it into triangles. 

PROBLEM VH. 

To find the area of a regular polygon. 

From (Geom., B. IV., th. 16), we derive the following 

RULE. 

Midtiply half the perimeter of the polygon by the perpendic- 
ular let fall from the centre on one of its sides, and the product 
will be the area. 

If the perpendicular be not given, it may be obtained in 
the following manner : 

Divide 360 degrees into twice as many equal parts as 
there are sides in the polygon ; then each one of these parts 
will be equal to half the angle subtended by a side. 

Now, half of one of the sides of the polygon, the perpen- 
dicular, and the line drawn from the centre to the extrem- 
ity of a side, constitute the base, perpendicular, and hypoth- 
enuse of a right-angled triangle, of which we have the base 
equal to half the side and the angles, from which the per- 
pendicular can be determined. 

EXAMPLE. 

Find the area of a regular pentagon the side of which is 
25 feet. 

Here — - -=36°=the vertical angle of the right-angled 

triangle. 

Then (Prop. 1, Plane Trig.), 

P 
tan. A= 7 ; 
o 

or p — b tan. A, in which p— perpendic- 

ular, and J = half side=12.5 feet; and A=90°— 360=54°. 
Putting it into logarithms, 

log. p — log. 5+ log. tan. A— 10. 

5=12.5 log. 1.096910 

A=54° tan. 10.138739 

^==17.205 1.235649 

.-. |(125 x 17.205)= 1075.3125 sq. feet, Ans. 



4 


u 


1.000000 


5 


it 


1.720477 


6 


u 


2.598076 


7 


a 


3.633912 


8 


a 


4.828427 


9 


a 


6.181824 


10 


a 


7.694208 


11 


a 


9.365639 


12 


a 


11.196152 



180 MENSURATION. 

If we take the side of a regular polygon equal unity, then, 
by the above formulas, we shall find the area of a 

Triangle, equilateral, 3 sides, 0.433012 

Square, 

Pentagon, 

Hexagon, 

Heptagon, 

Octagon, 

Nonagon, 

Decagon, 

Undecagon 

Dodecagon 

Now the areas of similar polygons are to each other as the 
squares of their like sides (Geom., B. IV., th. 11). Hence, 
to find the area of any regular polygon, we shall have the 
following 

RULE. 

Multiply the square of the side of the ■polygon by the tabular 
area set opposite the polygon of the same number of sides. 
Thus: 

Ex. Find the area of a regular decagon whose side is 10 
ieet. 

Tab. area=7.694208, and 10 2 =zl00 ; 
.♦. 7.694208 x 100=769.4208 sq. feet, Ans. 

PROBLEM VIII. 

Given the diameter of a circle to find the circumference ; or 

the circumference to find the diameter. 

From (Geom., B. IV., th. 17), we have the following: 

cir. = 27rr 

cir 
r~— , where 7r=3.1416 nearly; 

Ztt 

cir 
/. 2r=— =r diameter. 

7T 

Hence, multiply the diameter by 3.1416, and the product will 
be the circumference; or, divide the circumference by 3.1416, 
and the quotient will be the diameter. 



MENSURATION. 181 



PROBLEM IX. 



To find the length of an arc of a circle containing any num- 
ber of degrees. 

From (Plane Trig., Chap. I., sec. 4), we derive the fol- 
lowing 

RULE. 

Multiply the number of degrees in the given arc by the ra- 
dius of the circle, and divide the product by 57.29578. 

If the arc contains degrees and minutes, reduce the min- 
utes to the decimal of a degree. 

EXAMPLES. 

1. Required the length of the arc whose radius is 9, and 
the number of degrees 38° 56' 6". Ans. 6.11701. 

2. Required the length of an arc of 60°, the radius be- 
ing 10. 

PROBLEM X. 

To find the area of a circle. 

From (Geom., B. IV., th. 17), we have the following 

RULE. 

Square the radius, and multiply by 3.1416 ; 
Or, Square the diameter, and multiply by .7854. 

EXAMPLE. 

What is the area of a circle whose diameter is 7 ? 

Ans. 38.4846. 

PROBLEM XI. 

To find the area of a sector of a circle. 

Let abcO be any sector of a circle. 

Put Oa=Oc=r; 

And the angle aOc=6. 

180° 
Then the arc abc=rd-. (Trig., Chap. I.). 

7T 

Now, if the sector were a quadrant, the arc 
ale would be equal to \irr, and the area of the quadrant 
would be z=^7rr 2 . 




182 MENSURATION. 

But, in equal circles, sectors are to each other as the arcs 
on which they stand. 

/. Area of the sector : area of quadrant : : 7^775 : i^r ; 



180° 
Trrd 
180° 
Trrd 

1W 
From which we derive the following 



or, Area of the sector : ^nr 2 : : 777^ : \txt. 



\ Area of sector =%r x =r 7^5. 



RULE. 

Multiply half the radius by the arc on which the sector stands. 

EXAMPLES. 

1. Find the area of a sector of a circle whose arc con- 
tains 18°, the diameter being 3 feet. 

Trrd ^ e 3*14159 x 1.5x18 

Area =i rX 180-o = ' 75x 180 = 

.353428875 sq. ft. 

2. Find the area of a sector of a circle whose arc con- 
tains 10°, the diameter being 10 feet. 

Arts. 2.1816 sq.ft. 

PROBLEM XII. 

To find the area of a segment of a circle. 
The area of the sector CAEB (by the 

Trrd 
-last problem) is equal to J^Xt^o ; 

loO 

And the area of the triangle CAB= 
\r 2 sin. 6 (by prob. 1). 

Therefore the area of the segment AEB, ^ 
being the difference between the sector and 




the triangle, is U?L-— £ sin. d\r 2 ; 



And the area of the segment AGB, which is greater than 
a semicircle, is the sum of the sector AGB and the triangle 
ACB. 

Hence, to determine the area of a segment of a circle, we 
have the following 



MENSURATION. 183 



RULE. 



Find the area of a sector which has the same arc as the seg- 
ment ; also the area of the triangle formed by the chord of the 
segment, and the radii of the sector. 

Then take the difference of these areas when the segment is 
less than a semicircle, and the sum ivhen it is greater. 



EXAMPLES. 

1. Find the area of a segment of a circle, the chord be- 
ing 12 feet, and radius 10 feet. Ans. 16.3504 sfc[. ft. 

2. Find the area of a segment of a circle whose arc con- 
tains 10°, the radius being 10 feet. Ans. 6.336 sq. ft. 

PROBLEM XIII. 

To find the area of a circular ring ; or the area included be- 
tween the circumferences of two circles which have a common 
centre. 

Let R denote the radius of the greater circle. 

Let r denote the radius of the less circle. 

Then (Geom., B. IV., th. 17) the area of the larger 
is 7rK 2 , and the area of the less 

is 7rr 2 . 

/. 7rR 2 — 7rr 2 = area of the circular ring ; 
or, tt(R 2 — r 2 ), or tt(R + r ) (R — r). 

Whence we derive the following 

RULE. 

Multiply the sum of the two radii by their difference, and 
that product by 3.1416. 

EXAMPLE. 

What is the area of the space between two concentric 
circles whose radii are 10 and 6 ? Ans. 201.0624. 



MENSURATION OF SOLIDS. 



The mensuration of solids consists of two parts : 
1°. The mensuration of their surfaces; 
2°. The mensuration of their volumes. 

SUKFACES OF SOLIDS. 



PROBLEM I. 

To find the convex surface of a right prism. 

Let ABCDE— K be a right prism. 

Then, since the planes FE, FB are per- 
pendicular to the base, their line of inter- 
section FA is also perpendicular to the base 
(Geom., B. V., Prop. XVIL), and there- 
fore the angle FAB is a right angle. 
Hence the area of ABGF is=ABxAF. 
Similarly for all the other parallelograms. A 
Hence the area of the parallelograms tak- 
en together is 

AF X (AB+BC+CD+DE+EA) 

From which we derive the following 




RULE. 



Multiply the perimeter of the base by the altitude, and the 
product will be the convex surface. 

When the entire surface is required, we must add the 
area of the two bases to the convex surface. 



EXAMPLE. 



What is the convex surface of a cube, the length of each 
side being 20 feet? Ans. 2400 sq. feet. 



MENSURATION. 18i 



PROBLEM II. 

To find the convex surface of a right cylinder. 
Since the cylinder is the limit of the convex surface of 
the inscribed prism, whatever may be the number of its 
sides, therefore the convex surface of a cylinder is obtained by 
multiplying the circumference of its base by the altitude. 

EXAMPLE. 

What is the convex surface of a cylinder whose length is 
20 feet, and the circumference of its base 3 feet ? 

Ans. 60 sq. feet. 

PROBLEM III. 

To find the convex surface of a right cone, or right pyramid. 

Let BCDE — A be any right cone, 
BCDE its circular base. 

In the circle inscribe any regular pol- 
ygon, BCDEFG, and join AC, AD. . . . 
Then will these be the edges of the in- 
scribed pyramid. 

Bisect any side, CD in H ; join AH ; 
then AH is perpendicular to CD (Geom., 
B. I., th. 3, Cor. 1) ; and therefore the 
area of the triangle ACD=^CD x AH. G 

Now the line joining A with the point of bisection of any 
other side of the polygon is equal to AH. 

Hence the convex surface of the inscribed pyramid is 
=4 AH (BC + CD+DE+, &c), 
= |AH x perimeter of the base. 

Now this is true whatever be the number of sides of the 
polygon forming the base, and, hence, is true in the limit. 
If, therefore, we increase the number of the sides indefinite- 
ly, the polygon will become the circumference of the circle, 
and AH the slant height of the cone. Therefore, 

The convex surface of a right pyramid or a right cone is 
equal to half the slant height multiplied by the perimeter, or cir- 
cumference of its base. 

Rem. If we suppose the surface of the cone to be un- 




186 



MENSURATION. 



wrapped, it is evident that its surface will be a sector of a cir- 
cle whose radius is the slant height of the cone, and the arc 
of the same length as the circumference of the base of the 
cone. It is plain that the area of this sector is, as it should 
be, the same as that of the convex surface of the cone. 

PROBLEM IV. 

To find the convex surface of a frustum of a right pyramid. 

Since the section is similar to the base (Geom., B. VII., 
Prop. VII.), and the base is a regular polygon, it follows 
that the sides of the section are all equal to each other. 
Hence the convex surface is composed of equal trapezoids, 
whose common altitude is equal to the slant height of the 
frustum. But the area of any one of these trapezoids is 
equal to half the sum of the parallel sides multiplied by the 
altitude (Prob. VI.). Hence the convex surface of a frustum 
of a right pyramid is obtained by multiplying half the sum of 
the perimeters of the two bases by the slant height of the frustum. 

EXAMPLE. 

How many square feet are in the surface of a frustum of 
a square pyramid whose slant height is 10 feet ; also each 
side ^f the greater base is 3 feet 4 inches, and each side of 
the less base is 2 feet 2 inches ? Ans. 110 sq. feet. 




PROBLEM v. 

To find the convex surface of the 
frustum of a right cone. 

Let £BEc be a 
frustum of a right 
cone, and suppose 
a sector, OPQ, of ° 
a circle to be de- 
scribed with a ra- 
dius OP = AB ; 
^ and if its base PQ= circumference 
BE, we have seen that the area of the 
sector OPQ is the same as the convex 




surface of the cone ABE (Prob. III., Rem.). 



MENSURATION. 187 

Take Op— Ah, and describe the arc pq; then, as before, 
the area of Opq= convex surface of the cone Abe; and, 
therefore, the convex surface of the cone is equal to pqQP. 
Let the angle POQ=0. 

OP=OQ=R. 
Op = Oq —r. 

180 c 

7T( 



Then the area of POQ=:iR 2 x n -^5, and 



pOq=%r 2 x 



180 



o* 



.-. the area of 2Q p i?=^f^+ j^J(R-r) ; 

but yqttq is equal to the circumference of the lower base ; 
180 

7~Vf) 

-^-q= circumference of the upper base (Prob. XL, Mens. 

Surfaces) ; and R—? , = slant height of the frustum. There- 
fore, to find the convex surface of a frustum of a cone, we 
have the following 

RULE. 

Multiply half the sum of the circumferences of the two bases 
by the slant height 

Cor. If we bisect pP in t f , and describe the arc t't, then 
arc tt'^m+r}^s- 

Therefore the convex surface of a frustum of a cone is 

equal to the product of the slant Jieight, and the circumference 
of a section at equal distances between the two bases. 

EXAMPLE. 

What is the convex surface of a cone whose slant height 
is 10 feet, and the circumferences of its bases are 6 feet 
and 4 feet ? Ans. 50 sq. feet. 

PROBLEM VI. 

To find the surface of a spherical zone, and also that of a 
re. 




188 MENSURATION. 

Let AB be a quadrant of a circle whose 
radius is OA or OB. If we suppose the 
quadrant to revolve around AO it will 
describe a hemisphere ; and if we suppose 
a number of equal chords, Ap'jp'j/', &c, 
to be drawn from point to point of AB, 
these chords, in the revolution, will de- 
scribe frustums of cones. Now the arc 
AR is the limit of the chords Ap'-f- 
/>y+, &c, and hence the surface of the 
portion of the sphere described by AR 
will be the limit of the sum of the frus- 
tums of cones described by Ap r , p'p", &c. 

Let PQ be one of these chords ; draw PM, QN, perpen- 
dicular to AO. Draw Ot perpendicular to PQ, and tn 
perpendicular to AO. Now tn is the radius of the mean 
section of the frustum of the cone described by PQ, and 
therefore 

the area of that frustum = 2tt . tn x PQ (Prob. V.). 

Now to =0*. sin. 20A, and 

MN=PQ.sin. PQN; or, since PQN=zOA, 
MN=PQ.sin. ZOA; 
/. toxPQ=0*xMN. 
.*. surface of the frustum = 2?r x Ot x MN. 

If we draw pV, /> // n // , &c, perpendicular to AO, we 
shall have, for the sum of the frustums of the cones, 
2rrxOt (Aw'-f-nV' + , &c.) 
= 27r x Otx AN, if we only consider the por- 
tion of the sphere described by AR. 

This is true whatever may be the number of chords, and 
is, therefore, true in the limit ; but in the limit 02 becomes 
the radius of the sphere ; therefore the surface of a portion 
of the sphere whose height is AN=27r x OA x AN. 

But 2tt x OA — circumference of a great circle of the 
sphere, and AN is the altitude of the zone. Therefore, 

To find the surface of a spherical zone, we have the fol- 
lowing 

RULE. 

Multiply the altitude of the zone by the circumference of a 



MENSURATION. 189 

great circle of the sphere, and the product will be the surface 
required. 

Again, if we take the whole spere, then AN becomes 
2AO ; therefore the 

2 

surface of a sphere =47rX AO . 
Hence, to find the circle of a sphere, we have the following 

RULE. 

Multiply four times the square of the radius by 3.1416. 

If we imagine a right cylinder to be described about a 
sphere, its concave surface will be equal to the circumfer- 
ence of a great circle multiplied by the diameter of the 
sphere. Hence, 

The surface of a sphere is equal to the convex surface of the 
circumscribing cylinder. 

EXAMPLE. 

What is the surface of a globe whose diameter is 17 
inches ? Ans. 6.305 sq. feet. 



VOLUME OF SOLIDS. 

PROBLEM I. 

To find the volume of a right prism or cylinder. 

RULE. 

Multiply the area of the base by the altitude of the prism or 
cylinder, the product will be the volume (Geom., B. VII., Prop. 
5, Cor. 2). 

EXAMPLE. 

Eequired the volume of a triangular prism whose length 
is 20 feet, and the sides of its triangular bases 3, 4, and 5 
feet respectively. Ans, 120 cubic feet. 



190 



MENSURATION. 



PROBLEM II. 

To find the volume of a pyramid or cone. 

RULE. 

Multiply the area of the base by one third of the altitude of 
the pyramid or command the product will be the volume (Geom., 
B. VIL, Prop. 10, Cor. 1 and 2). 

EXAMPLE. 

What is the volume of a cone, the radius of its base be- 
ing 9 inches, and its height 15 feet ? 

Ans. 8.8357 cubic feet. 



PROBLEM III. 

To find the volume of a frustum of a right cone 

Let a5BA be a frustum of the cone 
ABC. Join COH where O and H are 
the centres of the bases of the frustums. 

Let AO = r; aH = R ; OH = h, and 
CO=z; 



Then, vol. CAB — J7rr 



") by Prob. 



vol. CEF=i7rR 2 (z+A)j II 
.-. vol. of frustum— ^7r{R 2 {x-\-h) — r 2 x} 

= ^7r.'R 2 h-j-^7r.(R 2 -r 2 )x, 
But, by similar triangles, 

x-\-h: R::;r : r ; 
or, h : x\\ R— r : r. 

hr 
•'• X= ~R^r 



Hence, vol. of frustum =+7r.R 2 /*-}- -J 7r(R 2 —r 2 ) 




hr 



R— r 

=^7r/z(R 2 4-Rr+r 2 ). 
From which we perceive that the volume of the frustum of 
a cone is equal to the sum of the volumes of three cones, whose 
common altitude is the altitude of the frustum, and whose bases 
are the lower base, the upper base, and a mean proportional to 
the two. From the above formula we also derive the fol- 
lowing 



MENSURATION. 191 



RULE. 



To the sum of the squares of the radii of the two bases add 
the product of these two radii; multiply the sum by one third 
the altitude, and that product by 3.1416. 

EXAMPLE. 

What is the volume of the frustum of a cone, the diame- 
ter of the greater end being 5 feet, that of the less end 3 
feet, and the altitude 9 feet ? Ans. 115.4538 cubic feet. 

To find the volume of a frustum of a regular pyramid. 

Add together the areas of the two bases and a mean propor- 
tional between them ; multiply this sum by one third of the alti- 
tude. 

EXAMPLES. 

1. What is the volume of the frustum of a cone, the di- 
ameter of the greater end being 5 feet, that of the less end 
3 feet, and the altitude 9 feet 1 

Ans. 115.4538 cubic feet. 

2. What is the volume of the frustum of a square pyra- 
mid, the larger end being 4 feet each way, the smaller end 
3 feet, and the height 6 feet ? Ans. 74 cubic feet. 

PROBLEM IV. 

To find the volume of a sphere. 

Since every sphere is two thirds of the circumscribing 
cylinder (Geom., B. VII., Prop. 12), and the volume of the 
circumscribing cylinder 

= Trr 2 x 2r (Prob. I. Vol . Solids) ; 
.*. vol. cir. cyl. = 27rr 3 . 
Or, if we put 2 r—T) —diameter/. r=JD, and 
vol. cir. cyl. = j7rD 3 ; 
/. vol. of sphere =j7r.D 3 . 
Hence we have the following 

RULE. 

Cube the diameter of the sphere, and midtiply by \ir» 
^=0.5236. 



192 MENSURATION. 



EXAMPLE. 

What is the volume of a sphere whose diameter is 6 
feet ? Ans. 113.0976 cubic feet. 

PROBLEM V. 

To find the volume of a spherical segment or zone. 

Let a = area of a great circle of the sphere. 

d=z diameter. 

^—altitude of zone or segment. 
Then (Geom., B. VII., Prop. 12, Cor. 3), 

vol. of segmental .~{M— 2h)h\ 

CL 

But a—\itd 2 . 

.*. vol. of segment =^tt(M— 2 h)h 2 . 
From which we derive the following 

RULE. 

From three times the diameter of the sphere subtract twice 
the altitude of the segment ; multiply the remainder by the square 
of the altitude, and this last product by 0.5236 for the volume 
required. 

Rem. 1. If the segment has two bases, and both on the 
same side of a plane passing through the equator, we may 
find the volume, by the above rule, under the supposition 
that we have two segments ; then take the difference of the 
volumes for the volume of the segment. 

Rem. 2. If the bases lie on different sides of the equator, 
take the sum of the volumes from the volume of the sphere 
for the volume of the segment. 

The formula for Rem. 1 is, 

vol.seg.=i7r{3^ 2 -/z /2 )-2(A 3 -^ 3 )J. 

And the formula for Rem. 2 is, 

vol. seg. =i7r{d 3 -3d(h 2 + h' 2 )-2(h*-{-h' 3 )i . 

EXAMPLE. 

In a sphere whose diameter is 21 inches, what is the vol- 
ume of a segment whose height is 4.5 inches'? 

Aiis. 572.5566 cubic inches. 




MENSURATION. 



PROBLEM VI. 



To find the volume of a wedge. 
is a solid, 




193 



Def. A we 
bounded by five planes ; the 
rectangular base, ABCD, two 
trapezoids, ABFE, DGCFE, 
intersecting in the edge EF, 
and two triangular ends, 
ADE, BCF. 

The altitude of the wedge is 
the perpendicular drawn from 
any point in the edge to the plane of the base, as EH. 

Put I— ABz=length of the base. 

r=EF= " « edge. 
£=BC:=AD=breadth of the base. 
7i= HE = altitude of the wedge. 

Now if the length of the base and that of the edge are 
equal to each other, it is plain that the wedge is half a par- 
allelopiped of the same base and altitude. 

If the length of the base is greater than that of the edge, 
let a plane, EG-I, be drawn parallel to BCF. The wedge 
will then be divided into two parts, viz., the triangular 
prism BCG-— F, and the pyramid AIGD— E. 

The vol. of prism=£W (Vol. of Solids, Prob. I.) 

vol. of pyr. -\bh(l-V), (Vol. of Solids, Prob. II.) 

/. vol. of wedge=$&/z(2J-f-0. 
If the length of the base is less than that of the edge, 
then the wedge will be equal to the difference between the 
prism and pyramid. 

.-. vol. of wedge =z±bh(2l+l f ), 
the same as above. 



Hence we derive the following 



RULE. 



To twice the length of the base add the length of the edge, 
and multiply the sum by one sixth of the product of the height 
of the wedge and breadth of the base. 



194 MENSURATION. 



EXAMPLE. 




What is the volume of a wedge whose base is 30 inches 
long and 5 inches broad, its altitude 12 inches, and the 
length of the edge 24 inches f Ans. 840 cubic inches. 

PROBLEM VII. 

To find the volume of a rectangular prismoid. 

Def A rectangular prismoid is a solid 
bounded by six planes, of which the two 
bases are rectangles, having their corre- 
sponding sides parallel, and the four oth- 
er sides are trapezoids. 
Put Z r r=AD=:BC; 
b'=AB=CT>; 
Z=HE = GF; 

•7z=alt. of the prismoid ; 
M=the length of the mid section =pq; 
??z=the breadth " " =pr. 

If a plane be passed through the opposite edges of the 
upper and lower bases, the prismoid will be divided into 
two wedges, whose bases are the bases of the prismoid, and 
whose edges are V and I. 

The volume of these wedges, by Prob. VI., is 

=ih(2bl-\-bl' + 2b'l' -\-b' 7)= vol. of the prismoid. 
But, since M and m are the length and breadth of the 
mid sections, we have 

2M=Z+r, and 2m=b+b' 
;AMmz=(l+l'){b+b')=zbl+bl'+b'l+b'l'. 
Substituting 4Mm for its value in the preceding expres- 
sion, we have for the value of the prismoid 

P(M+4Mm+&7'), 
from which we derive the following 

rule. 

To the sum of the areas of the two bases add four times the 
area of a middle section, and multiply the sum by one sixth of 
the altitude. 



MENSURATION. 



195 



EXAMPLE. 



What is the volume of a block of marble in the form of 
a rectangular prismoid, the length and breadth of one end 
being 16 and 12 inches, and of the other 7 and 4 inches, 
the length of the block being 24 feet 1 

Ans. 16^ cubit feet. 



PROBLEM VIII. 

To determine the volume of a railroad cutting. 

In Prob. VII. the figure is very nearly that of a portion 
of a railroad cutting in which AC is the road, and AG 
and DF the sloping sides of the embankment ; hence the 
solid contents required can be found by means of the rule 
given in the last problem. 

It is to be observed that the rule requires that HG and 
EF be straight lines, or, as an approximation, to be very 
nearly straight lines ; or AG- to be a plane, which is not 
true if the cutting is a long one. For this reason we shall 
derive the following : 



Let ABCD represent a 
section, made by a vertical 
plane, of the hill to be cut 
through ; AB the level of 
the road ; and suppose sec- 
tions of the cutting to be 
made by planes perpendic- 



■SJS-B 




PW^^U 



ular to AB, at equal distances along that line, viz., at Mj, 

M 2 M 2re+1 . 

Let the terminal sections at A and B be denoted by a 
and b respectively, and the sections at P 1? M 3 , P 2 , M 2 . . . . , 

&c, P 2n+] , M 2fl+1 , be denoted by p x ,p 2 jp 9n+1 , the number 

of sections being odd. And let the common distance between 
the sections equal h. Then (by Prob. VII.), 

vol. of AM2P 2 Drr:^AM 2 («+4pi-f^ 2 ); or 

=iHa+4pi+P*) > and 
vol. of P 2 M 2 M 4 P 4 =-P(p 2 +4p 3 -f j9 4 ), and so on ; 



vol. ofP 2n M 2n BC: 



iHP, n + ^n+l)+0. 



Hence, by addition, the whole required volume 



196 MENSURATION. 

=i*{(«+*)+4(Pi-ri*+> &c /Wi)+ 2 0>a+lVf- 

"Whence we derive the following 

RULE. 

Between the first and last sections, make an odd number of 
sections at equal distances along the road, the planes of the sec- 
tions being perpendicular to the road; then one third part of the 
common distance, multiplied by the sum of the first and last sec- 
tions, with four times the sum of the odd sections and twice the 
sum of the even sections, gives the volume required. 



SPHERICAL TRIGONOMETRY. 



CHAPTER I. 

1. Spherical Trigonometry treats of the relations sub- 
sisting between the parts of a solid angle formed by the 
inclination of three planes ; or, supposing the angular point 
the centre of a sphere, the relations between the sides and 
angles of the triangular surface of the sphere inclosed by 
the three planes. 

2. A spherical triangle is the portion of the surface of a 
sphere inclosed by the three planes which form the solid 
angle at the centre : it is bounded by three arcs of great 
circles. 

3. A spherical angle is the angle contained by any two of 
the planes passing through the centre of the sphere. 

4. The sides of a spherical triangle are the three arcs 
which measure the rectilinear angles at the centre respect- 
ively. The sense in which the spherical triangle is em- 
ployed being once understood, we may transfer our atten- 
tion from the solid angle to the triangle, and the solid angle 
need not be represented in our diagrams. 

The first formula which we will establish is to spherical 
trigonometry what the expression for the sine of two angles 
is to plane trigonometry ; that is, it serves as a key to all 
the rest. 

PROBLEM I. 

To express the cosine of an angle of a spherical triangle in terms 
of the sines and cosines of the sides. 
Let ABC be a spherical triangle, O the 
centre of the sphere. Let the angles of the 
triangle be denoted by the large letters A, 
B, C, and the sides opposite to them by the 
corresponding small letters «, b, c. At the 
point A draw AT, a tangent to the arc 
AB ; and AT 7 a tangent to the arc AC. 




198- SPHERICAL TRIGONOMETRY. 

Then the spherical angle A is equal to the angle TAT 7 
between the tangents, because the tangents AT, AT 7 are 
drawn perpendicularly to the radius of the sphere at the 
point A. Therefore the angle TAT 7 is equal to the angle 
made by two planes passing through AB, AC ; that is, to 
the spherical angle TAT 7 . 

Join OB, and produce it to meet AT in T. 

Join OC, and produce it to meet AT in T 7 ; join TT 7 ; 

OT 

then — — —sec. AB =sec. c ; 

OC 

OT 7 

— — -=sec. AC =:sec. b ; 
OC 

AT 

— — =tan. AB=tan. c ; 

OC 

AT 7 

—tan. AC = tan. b. 
OC 

Then, in the triangle TOT 7 , we have 

TT 72 =OT 2 +OT 72 -20T.OT 7 cos^TOT 7 . 

Divide each member of the equation by OC , we have 

TT /2 OT 2 OT /2 OT OT 7 

L + !±L^_2^. ^L cos. TOT 7 



OC 2 ~OC 2 ' OC 2 OC OC 

=■ sec. 2 c -f sec. 2 b — 2 sec. c sec. b cos. a . (a) 
Again, in the triangle TAT 7 , we have 

TT ?2 =AT 2 +AT 72 -2AT. AT 7 cos. TAT 7 . 
Divide each member of this equation by OC 2 , it becomes 

TT 72 _AT 2 AT- AT AT 

0c 5 - oc 54 " oc 2 "" oc* oc cos * x 

=tan. 2 c-ftan. 2 b— 2 tan. c tan. b cos. A (b) 
Equating (a) and (b), we have 

tan. 2 c -f- tan. 2 6— 2 tan.c tan. b cos. A=sec. 2 c+sec. 2 &— 

2 sec. c sec. b cos. a ; 

sec. 2 c— tan. 2 c + sec. 2 J— tan. 2 6=2 sec. 6 sec. ccos.a— 2 tan. b 

tan. c cos. a. 
But sec. 2 c— tan. 2 c=l, and sec. 2 b— tan. 2 b—1. 

Substituting these in the above equation, it becomes 
2r=2 sec. b sec. c cos. a— 2 tan. b tan. c cos. A ; 
or tan. b tan. c cos. A=sec. b sec. c cos. a — 1. 



SPHERICAL TRIGONOMETRY. 199 

TT sin. b sin. c cos. A cos. a 
Hence = = 1 

COS. COS. C COS.0 cos. c 

cos. a— cos. b cos. c 
~ cos. b cos. c 

cos. a— cos. J cos. c 

therefore, cos. A= ■ — = — : . . . (1) 

sin. o sin. c 

In like manner we find 

*, cos. b— cos. a cos. c ~, 

cos.B= : : ... (2) 

sin. a sin. c 

• cos. c— cos. a cos. J 

COS. C= : ; = ... (3) 

sin. a sin. 



PROBLEM II. 

Given tivo sides and the included angle of a spherical triangle, 
to determine the third side, independently of the unknown 
angles. 

Let b, c, A be given, and let it be required to find a, in- 
dependently of the angles B and C. 
Solution. Resuming formula (1), 

cos. a— cos. # cos. c 

cos. A= -. — = — : , 

sin. b sin. c 

we have cos. «=cos. A sin. b sin. c-f-cos. b cos. c. 

In order to adapt this equation to logarithmic computa- 
tion, let us add and subtract sin. b sin. c to the second mem- 
ber of the equation ; then 

cos. a=cos. A sin. b sin. c-f-cos. b cos.c+sin. b sin. c — 
sin. b sin. c ; 
=cos. b cos. c-fsin. b sin. c-fsin. b sin. c cos. A— 

sin. b sin. c ; 
=cos. (b— c)— sin. b sin. c versin. A. 
And by subtracting each member of this equation from 
unity, we have 

1— cos. a=l — cos. (b— c)-f-sin. b sin. c versin. A ; 
/. versin. a=versin. (b— c)-f sin. b sin. c versin. A 



. ,_ f sin. b sin. c versin. A") 

: versin. (5— c) < 1-j r — 77 r r 

C versin. (0 — c) ) 



versin. (b— c) 

^ , . _ sin. b sin. c versin. A 

Put tan. 2 6— : — jz : 

versin. (b—c) 



200 SPHERICAL TRIGONOMETRY. 

then versin. a=versin. (b — c)(l -f tan. 2 6) ; 

=versin. (b — c) sec. 2 . . . (4) 
Whence a is determined. 

PROBLEM III. 

Given two sides and the angle opposite to one of them, to de- 
termine the third side, independently of the angle opposite to 
the other side. 
Solution. Let a, b, A be given, to find c independently 

of the angle B. 

_. . . cos. a — cos. b cos. c 
Equation (1) : cos. A= : — -— ; 

sin. b sin. c 

:. sin. c sin. b cos. Az=cos. a— cos. b cos. c ; 

7 . cos. a 

or sin. c tan. b cos. A= =■ — cos. c. 

cos. b 

_ „ sin. 

Put tan. 0=: -=tan. b cos. A; 

cos. a 

sin. 6 cos. a 

then sin. c j. 4- cos. c = =- : 

cos. v cos. 6 

. _ _ cos. a cos. 

or sin. c sin. 6^+ cos. c cos. 6/= = — ; 

cos. b 

cos. a cos. 

/. cos. (c— 6)— — . . . (5) 

7 cos. b 

Whence c may be found, 6 being previously determined from 

the equation tan. 0=tan. b cos. A. 

PROBLEM IV. 

To express the cosine of a spherical triangle in terms of the 
sines and cosines of the angles. 
Solution. Let A, B, C, a, b, c be the angles and sides of 
a spherical triangle ; A/, B', C', a', V, c / the corresponding 
quantities in the polar triangle ; then, by (1), 

cos. A/ 



sin. V sin. & 
But (by Spherical Geometry, Prop. 6) 
A / =(180°—a), a / =:(180 o -A), &'=(180°— B), c'= 

(180° -C). 
.-.cos. (180°— a) = 
cos. (180°— A)— cos. (180°— B) cos. (180°— C) m 
sin. (180°-B) sin. (180°-C) ; 



SPHERICAL TRIGONOMETRY. 201 

cos. A -f- cos. B cos. C ■■ 

:. cos. a= : — ^— - — p; ... (6) 

sin. B sin. C 

In like manner we obtain 

, cos. B4-cos. A cos. C .„ 

cos. b = . . — ... (7) 

sin. A sin. C 

cos. C-f-cos. A cos. B ._. 

cos. c= : t — ^^5 . . . (8) 

sin. A sin B v 

PROBLEM V. 

Given two angles and the included side, to find the remaining 

angle, independently of the unknown sides. 

Solution. Let A, B, c be given, to find C independently 

of a and b. 

_, ' cos. C+cos. A cos. B 

By (8), cos. c— : -r—. — ; 

v sin. A sin. B 

.*. cos. C=cos. c sin A sin. B— cos. A cos. B. 
Subtracting each member of this equation from unity, we 
have 
1— cos. C = l— sin. A sin. B (1— versin. c)-f cos. A cos. B ; 

= l + cos. (A-f B) + sin. A sin. B versin. c; or 
2 sin. 2 -^C = 2 cos. 2 i(A-|-B)-}-sin. A sin. B versin. c; 

» w * ™ f ^ sm - A sin. B versin. c") 
=2 co,H(A+B) |l+ 2cos ,, (A+B) }• 

_ » _ sin. A sin. B versin. c 

Let tan. 6= — — — — — ; 

2cos. 2 -|(A+B) 

then sin. 2 -JC=:cos.H(A+B) sec. 2 (9 ... (9) 

PROBLEM VI. 

Given two angles and the side opposite to one of them, to find 

the remaining angle. 

Let A, B, a be given to find C. 

Solution. 

_ ,„. cos. A+cos. B cos. C 

By (6), cos. a— : — =— : — ; 

J v " sin. B sm. C 

.-. sin. B sin. C cos. a=cos. A cos. B cos. C ; 

• r^ t-> cos. A _. 

or sm. C tan. B cos. a= —4- cos. C ; 

cos. B 

„ __ „ cos. A 

or sm. C tan. B cos. a— cos. C 



cos. B 
12 



202 SPHERICAL TRIGONOMETRY. 

Put tan. B cos. a=tan. 6. 



cos. 6 



. sin. 6 cos. A 

then sin. C t,— cos. C= - ; 

cos. 6 cos. B 

or sin. C sin. 6— cos. C cos. 0= — ^rcos. 6; 

cos. B 

4.x, e /rt . , m cos. A cos. 6 

therefore — cos. (04-6)= — . . (10) 

v ' cos. B v ' 

Whence C may be found, 6 being known from the equa- 
tion tan. 0=tan. B cos. a. 

problem vn. 

To express the sine of an angle of a spJierical triangle in terms 

of the sines of the sides of the triangle. 

_, /<x . cos. a — cos. b cos. c 

By (1), cos. A= : — ; — : . 

sin. b sm. c 

By adding unity to both members of the equation, we 

have 

• , . cos. a— cos. b cos. c+sin. b sin. e 

1-j-cos. A= : — - — r— ' ; 

sin. b sm. c 

cos. a— cos. (b+c) 



sin. b sin. c 

2 sin. -J(a+£+c) sin. \{b-\-c— a) 

sin. b sin. c 

Put s=-^(a+5+c) ; then s— a—^(b-{-c— a) ; s—b= 

^{a-\-c— b); s—c=^(a-\-b—c); 

2 sin. s sin. (s— a) , . 

.\l+cos.A= : — t-t 1 . . (a) 

sm. b sm. c 

Again, resuming the expression for cos. A, and subtract- 
ing each member from unity, we shall have 

cos. b cos. c+sin. b sin. c— cos. a 

1— cos. A= : — z—. ; 

sm. b sm. c 

cos. (b— c) — cos. a 

sin. b sin. c 

2 sin.-^(a+&— c) sin.J(a+c— b) 

sin. b sin. c 

2 sin. (s—c) sin. (s—b) 

sin. b sin. c 



<») 



SPHERICAL TRIGONOMETRY. 203 

Multiplying equations (a) and (b), we have 

4 sin. s sin. (s— a) sin. (s— b) sin. (s—c) 

1— cos. 2 A= ; — 5-5 — : — 5 ; .*. 

sin/ 6 sin/c 



sin.A=- — — : Vsin.ssin.(s— a) sin. (5— 6)sin.(s— c) (11) 

sm. bsm.c 

In like manner, 

2 



sin. 13=:- ; V sin. s sin. (s— a)sin.(s— b)sm.(s— c) (12) 

sin. a sin. c 



2 



sin.C=:- : — 7 Vsin.ssin.(s — a)sin.(s— 5) sin. (5— c) (13) 

sin. a sin. J v ' ' v 

Also, by equation («), we have 

- , .2 sin. s sin. (s—a) 

1 + COS. A = : — — '- ; 

sm. sm. c 

_ _ _ . 2 sin. s sin. (s—a) 

or 2cos. 2 AA= : — — -h -. 

sin. b sm. c 



i a /sin. s sin. (s—a) 
/.cosiA=\/ — : — — J ' . . (14) 

4 V sin. b sm. c v ' 

In like manner, 



1X) /sin. s sin. (s— &) 
cos.£B=\/ — : p i . . (15) 

V sin. a sm. c 

, ri a /sin. s sin. (s—c) ,., „ 

cos.^C^W — : . v ' . . (16) 

V sin. a sm. 6 
Again, by equation (5), 

1 _cos. A= 2sm.(«-c)BM«-j) 

sin. & sin. c ' 

2 sin.' jA = 2sia - ( !-^ sin -<!^£>. 

sin. 6 sin. c 

Hence sin. jA^/^"-^ ^('-"A . ( i 7) 

V \ sin. b sin. c ) 

Also sin. i B = ./( sin -< S - a)sin - (s - C ) N ) 

V \ sm. a sm. c / 

-.^ //sin. (s—a) sin. (s — 5)\ 

sin.£C=\/ H — ^-r-4 ) ■ (19) 

2 V \ sm. a sin. 6 / J 

Again: if we divide equations (17), (18), (19), by equa- 
tions (14), (15), (16) respectively, we shall obtain, 
tan /sl^^njJ=7) _ 

V sm. s sm. (s—a). 



(18) 



!04 SPHERICAL TRIGONOMETRY. 



ta P .iB = v/ gin -.<'- a . )8i / n -('- c) . (21) 

V sin. s sin. f,s — /n v 



sin. s sin. (s — #) 



' /sin. (s— a) sin. (s— b) ._,__ 

tan.£C=\/ r^ : , v ' . (22) 

V sin. s sin. (s—c) v 



sin. s sin. (s—c) 

problem vm. 

To express the sine of a side of a spherical triangle in terms of 
the sines and cosines of the angles. 

Solution. By (6) : 

cos. A -4- cos, B cos. C 

cos. a— r— 5— — n * 

sin. B sin. C 

Add unity to both members, and we have 

cos. A+cos. B cos. C+sin. B sin. C 

1 +cos. a- : — -— — ; 

sin. B sin. C 

cos. A-j-cos. (B — C) 

sin. B sin. C 

_2 cos.i(A-fB-C) cos.£(A+C— B) 

sin. B sin. C 

Put/=J(A4-B+C); thena / -A=-i(B+C-A); s'- 

B=:i(A+C-B); s'_C=£(A+B-C); therefore, 

2 cos. (/— C) cos. (s'-B) , . 

1 + cos. a= V . J . ^ ' . . (a) 

sin. B sin. C v 

Resuming the expression for cos. a, and subtracting each 

member from 1, we have 

cos. B cos. C — sin. B sin. C-f-cos. A 

, 1— cos.a= , . _, ; 

sin. B sin. C 

cos. (B+C) + cos.A 



sin. B sin. C 
2cos.£(A+B+C)cos.j(B+C— A) ^ 

sin. B sin. C 
2 cos.s' cos. (s'—A) 

= sin. B sin. C * ' 

Multiply equations (a) and (b), and we shall have 

4 cos. s / cos. (/—A) cos. (/— B) cos. (s / — C) 

1 —cos. 2 a=z * . o -p . \ „ — ' " " 

sin. 2 B sin. 2 C 

therefore, 



SPHERICAL TRIGONOMETRY. 205 



2 



sin. a—- — — - — - x 
sin. B sin. C 

V— cos./ cos. (/—A) cos.(/— B) cos. (/— C) (23) 

In a similar manner, 

. , 2 

sm. b—- — : — - x 

sm. A sin. C 

V— cos./ cos. (/—A) cos. (/— B) cos. (/— C) (24) 

2 

Sin. C=- r r— =: X 

sm. A sin. B 

V— cos. / cos. (/—A) cos. (/— B) cos. (/— C) ( 25 ) 

By equation (a) we have 

2 cos. (/-B) cos. (/-C) 

l + COS. a — ^ i ^ ' ; 

sin. B sin. C 

2 cos. (/-B) cos. (/— C) 

sm. B sm. C 

. /cos. (/-B) cos. (/-C) .... 

.-. cos. \a—\ v . ' . v _, . . (26) 

2 V sm. B sin. C v ' 

ai it /cos. (/— A) cos. (s / — C) ,_>_. 

Also cos. U—\/ ^ — /— ; — ^- . (27) 

J V sm. A sm. C v 

cos Ic-a / C0S ' (/ ~ A) C ° S - (/ ~ B) (28^ 

cos.^_y sin . Asin . B • (<«) 

By equation (5), 

2 cos. / cos. (/—A) 

1 —cos. a— : — . v ; 

sm. B sm. C 

^ . ., —2 cos. /cos. (/—A) 

.-. 2 sin. 2 i«= . _ . v ^ ± ; 

2 sm. B sin. C 

. , /—cos. /cos. (/—A) /rt ^ v 

hence sin.-kz=\/ . p . ^ • - (29) 

2 V sin. B sm. C v ' 

17 /—cos. /cos. (/—B) /o ^ 

also sm.U=\/ : — A . v _. — '- . . (30) 

2 V sin. A sm. C v ' 



have 



. . /—cos. / cos. (/— C) ,„' 

sin.-|c=\/ : — .-I— ' . . (31) 

2 V sm. A sm. B v ' 

Again, by dividing formulas (29, 31) by (26 and 23), we 



i / —cos. /cos. (/—A) ,„. 

tan ^=V co,(/-B)J.( S --C) ' < 32 > 



206 SPHERICAL TRIGONOMETRY. 



tan i&_4 / -cos. ^ cos. (/-B) 
* 2 ~Vcos.(s'-AW.(/-( 



(«'— A)cos.(/-C) 



(33) 



. / — cos. /cos. (/—C) *. 

tan --^Vcos.(/-A)cos.y=lj • ( 84 > 
Although the formulas (23-25), (29-34), appear under 
an impossible form, they are, in reality, always possible. 

For, since the sum of the angles of a spherical triangle 
is always greater than two and less than six right angles 
(Spherical Geom., Prop. 9), therefore 

A+B+C>180°and<540°; 
therefore i(A+B+C)>90° and<270°; 
hence cos. s' is always negative, and — cos. / is always pos- 
itive. 

Again, if a / , b\ c' be the three sides of the polar triangle, 
then, because the sum of any two sides of a spherical trian- 
gle is greater than the other side, 

.-. 180°-B+180°-C>180°-A; 
hence B+C— A<180°, 

and i(B+C-A)<90°; 

consequently, cos. (/—A) is always positive. In like man- 
ner we can prove that cos. (/— B) and cos. (/— C) are al- 
ways positive. 

Hence the foregoing formulas are possible in eveiy case. 

PROPOSITION I. 

In every isosceles spherical triangle, the angles opposite the equal 
sides are equal. 

. cos. a— cos. b cos. c 
(1) cos. A= : — — ; 

v ' sin. b sin. c. 

_ cos. b— cos. a cos. c 

COS. B = : : . 

sin. a sin. c 

But if the triangle is isosceles, a=b; 

cos. a— cos. a cos. c 

.-. cos. A= : : ; 

sin. a sm. c 

_ cos. a— cos. a cos. c 

and cos. B = : : ; 

sm. a sm. c 

.\A=B. 



SPHERICAL TRIGONOMETRY. 207- 



PROPOSITION II. 

Conversely, if the angles of a spherical triangle are equal to 
each other, the triangle is isosceles. 

cos. A+cos. B cos. C 
(6) cos. a— : — =—. — ^ ; 

v J sin. B sin. C 

cos. B+cos. A cos. C 

COS. 0= ; t : 7^ • 

sin. A sin. L> 

But B=A; 

cos. A+cos. A cos. C 

.-, cos. a= : — t — ; — ~ ; 

sin. A sin. (J 

_ cos. A+cos. A cos. C 

.'. COS. I— ; r ; -^ ; 

sin. A sin. C 
• a=b. 



CHAPTER IT. 

THEOREM II. 

The sines of the angles of a spherical triangle are to each other 
as the sines of the sides opposite to them. 

Demonstration. 



B y (ll),(12), S in.A= 2VsiD " Sin - (5 ri Sin - (S ~' )Sin - 5 ~ C) 5 
J v yn y ' sm. & sin. c 

. ._ 2-i/sm.ssin.(s— <2)sin.(s — b)sm.(s— c) 

sm.B= — *—. — - — it- 1 * -. 

sm. a sm. c 



Dividing (11) by (12), we have 

sin. A sin. a sin. c sin. a 
sin. B~~sin. b sin. c~ sin. b 

In like manner we obtain 

sin. A sin. a sin. 6__sin. a 
sin. C~sin. c sin. b sin. c 
sin. B sin. b sin. a sin. b 
sin. C — sin. c sin. a sin. c 



(35) 

(36) 
(37) 



208 SPHERICAL TRIGONOMETRY. 



PROBLEM I. 

To express the tangent of the sum and difference of tivo angles 
of a spherical triangle in terms of the sides opposite to these 
angles, and the third angle of the triangle. 
Solution. By (1), (3) : 

cos. a — cos. b cos. c . . 

COS. A = ; = ; ....(«) 

sin. b sin. c 

_, cos. c — cos. a cos. b 

cos. C = : ; — 7 ; 

sin. a sin. b 

.\ cos. c=cos. a cos. &-|-sin. a sin. b cos. C . . (b) 

Substitute this value of cos. c in equation (1), we have, 

cos. a — cos. a cos. 2 b — cos. b sin. a sin. b cos. C 

cos. A=: : — - — : ; 

sin. b sin. c 

cos. a (1— cos. 2 b) — cos. b sin. a sin. b cos.C 

sin. b sin. c 

cos. a sin. b— cos. b sin. a cos. C . . 

= ; ...(c) 

sin. c v 

Also, by substituting the value of cos. c in equation (b), 

in the formula (2), we shall have 

_, cos. b sin. a— cos. a sin. b cos. C , ,. 

cos. 13 = : . . (d) 

sin. c v 

Adding equations (c) and (d), we have 

cos. A + cos. B= 

sin. a cos. b-\- sin. b cos. a — (sin. a cos. J-}- sin. b cos. a) cos. C 

sin. c 

sin. (a-\-h) — sin. (a-\-b) cos. C 

sin. c 

n /rt „. sin. A sin. a 

But, by (35), -r— ^=z- — 7 ; 

' ' v 7 sm. B sin. b 

whence, by the theory of proportions, 

sin. A d= sin. B sin. a ± sin. b 



sin. B sin. b 

. . . ^, / . . 7v sin. B 

.*. sin. A±sin. B=(sm. «±sm. b) — — - : 

sin. b 

=(sm.a±sin. b) — . (e) 

sm. c v ' 

Dividing eq. (e) by (d), and taking first the positive sign, 



SPHERICAL TRIGONOMETRY. 209 

sin. A+sin. B sin. a+sin. b srn ' C 
cos. A + cos. B~~ sin. (a+&) 1— cos. C' 
2 sin. K A +B) cos. 4(A— B) 



2 cos. M A + B ) cos. i(A— B) 
2 sin. J(a-[-5) cos. \{a — b) 



2 sin. J(a+i) cos. £(a+£) 



cot. AC. 



or 



•■• tan -^ A + B >=3Si cot4C - • (38) 

Again, by taking the negative sign, we shall have 
. sin. A — sin. B sin. a— sin. b sin. C 
cos. A -f- cos. B — sin. {a-\-b) ' 1— cos. C ' 
2 sin. K A ~ B) cos. K A + B) 
2 cos.^A— B) cos.|(A+B)~ 
2 sin. ^(a—b) cos. ^(a+#) sin. C 



2 sin. £(a + £) cos - i(«+*) " 1 — cos. C ' 

therefore tan.4(A-B)=^l|g=|j oot.JC . (39) 

Formulas (38) and (39) may be written thus: 

cos. \{a-\-b) tan. ^(A-f-B)=cos. ^(a—b) cot. -|C ; 
sin.-i(a+£) tan. £(A— B) — sin.-J(a— b) cot.-^C. 

That is, in any spherical triangle, as the cosine of half the sum 
of two sides is to the cosine of half their difference, so is the co- 
tangent of half the included angle to the tangent of half the sum 
of the other two angles. And 

As the sine of half the sum of two sides is to the sine of half 
their difference, so is the cotangent of half the included angle to 
the tangent of half the difference of the other two angles. 

This is the first of Napier's Analogies. 

In a manner precisely similar will be found 

tan. i( B+C)=^ig=£>cot. 4 A . (40) 

tan -( B - c )=I-£f-Si cot - A • < 41 > 

tan :4 (A + C)=^i^co, i B - (42) 
tan.MA-C^i^jcot.iB . (43) 



210 SPHERICAL TRIGONOMETRY. 



PROBLEM II. 

To express the tangent of the sum and difference of two sides 
of a spherical triangle in terms of the angles opposite to them 
and the third side of the triangle. 
Solution. Let A, B, C, a, b, c, be the angles and sides of 

a spherical triangle ; A', B 7 , C', a', b', c / , the corresponding 

parts of the polar triangle ; then, by (38), 

tan. i(A / +B / )= C ° S ' f ^"12 cot.£C 
2V ^ ' cos. -^ +6') 2 

.-. tan. \ 5(180°— a) + (180°-6)J = 
cos.ij(180Q-A)-(180O- B)| Hm o_ e) . 
cos.£{(180 o -A) + (180°-B)J C ° ' 2l U C) ' 



Again, by (39): 

tan. ±{(180°— a)-(180°—b)\ = 



_ Bm.j(gt-b / ) 
-sm.i(a'+b') COt * 2C ' 



sin.^(180Q-A)-(180°-B)j 
sin.i{(180o_A)+(180o_B)} ° 0t - 2{i ™ ~ C) ' 

.-. tan. i(a-b)= Sm ' |ffi~ S tan. & . . . (45) 

2V ' sin. £(A+B) 2 v ' 

In like manner we shall obtain 

tan -« J -^£t(fe§ tan > • (47) 

tan. i(a+C )^|^§tan^ . (48) 

Or, as Me cosine of half the sum of two angles is to the cosine 
of half their difference, so is the tangent of half the included side 
to the tangent of half the sum of the other two sides ; and 



SPHERICAL TRIGONOMETRY. 211 

As the sine of half the sum of two angles is to the sine of half 
their difference, so is the tangent of half the included side to 
the tangent of half the difference of those sides. 

This is the second of Napier's Analogies. 

PROBLEM HI. 

To express the cotangent of an angle of a spherical triangle in 
terms of the side opposite, one of the other sides, and the an- 
gle contained between these two sides. 

Solution. From (36), sin. c=- r sin. a; 

sin. A 

and from (eq. c, Prob. 1), 

cos. A sin. c=cos. a sin. b — sin. a cos. b cos. C ; 

hence cos. A . sin. a=cos. a sin. b— sin. a cos. b cos. C; 
sin. A 

.*. cot. A=cot. a sin. b cosec. C— cos. b cot. C . . (50) 

If we substitute the value of cos. b derived from equation 

(2) in equation (1), we shall find a value for cot. A in terms 

of a, c, B. 

Thus cot. A=cot. a sin. c cosec. B^-cos. c cot. B (51) 

Proceeding in like manner, we shall obtain for the other 

angles, 

cot. B==:cot. b sin. a cosec. C— cos. a cot. C (52) 

cot. Br=cot. b sin. c cosec. A— cos. c cot. A (53) 

cot. C = cot. c sin. a cosec. B— cos. a cot. B (54) 

cot. C=cot. c sin. b cosec. A— cos>£ cot. A (55) 

PROBLEM IV. 

To express the cotangent of a side of a spherical triangle in 
terms of the opposite angle, one of the other angles, and the 
side interjacent to these two angles. 

Solution. Using the same notation as before for the an- 
gles and sides of the spherical and polar triangles, we have, 

by (50), 

cot. A'=cot. a' sin. V cosec. C / — cos. V cot. C 7 ; 
.-.cot. (180°— a)=cot. (180°— A) sin. (180°— B) cosec. 
(180°-c)-cos. (180°— B) cot. (180°— c) ; 
or —cot. a= —cot. A sin. B cosec. c— cos. B cot. c ; 
.*. cot. a=cot. A sin. B cosec. c+cos. B cot. c (56) 



212 SPHERICAL TRIGONOMETRY. 

Applying the same process to the other formulas in Prob. 
III., we shall obtain analogous results. Thus 

cot. a=cot. A sin. C cosec. 6 -[-cos. C cot. b (57) 

cot. £=cot. B sin. A cosec. c+cos. A cot. c (58) 

r=cot. B sin. C cosec. a+cos. C cot. a (59) 

cot. c=cot. C sin. A cosec. 5-fcos. A cot. c (60) 

= cot. C sin. B cosec. a+cos. B cot. a (61) 



EIGHT-ANGLED SPHERICAL TRIANGLES. 
CHAPTER III. 

SPHERICAL TRIANGLES OF ONE RIGHT ANGLE ONLY. 

A spherical triangle consists of six parts ; any three of 
these being given, the rest may be found. In the present 
case, one of the angles is supposed to be a right angle. If 
any other two parts be given, the other three may be de- 
termined. 

Now the combination of five quantities, taken three and 
three, is equal to 

5.4.3 , rt 

nor 10 - 

Therefore ten different cases present themselves in the so- 
lution of right-angled spherical triangles. 

Two rules have been invented, by the help of which we 
are enabled to solve every case of right-angled spherical tri- 
angles. They are known by the name of Napier's Rules for 
Circular Parts, and are as follows : 

The right angle is altogether thrown out of consideration. 
The two sides, the complements of the two angles, and the 
complement of the hypothenuse, are called the circular parts. 
Any one of these circular parts may be called a middle part, 
and then the two circular parts immediately adjacent to the 
right and left of the middle part are called adjacent parts ; 
the other two remaining circular parts, each separated from 
the middle part by an adjacent part, are called opposite parts. 
This being stated, we may now give the 



SPHERICAL TRIGONOMETRY. 213 



RULES. 

1. M being the middle part, the product of sin. M and tab- 
ular radius is equal to the product of the tangents of the adja- 
cent parts. 

2. The product of sin. M and tabular radius is equal to the 
product of the cosines of the opposite parts. 

In order to make these rules clearly understood, we will 
show the manner in which they are applied in various 
cases. It will assist the student to remember these, rules 
by observing that the first syllable of tangent and adjacent 
each has the letter a in it, and cosines and opposite the let- 
ter o. 

Let A, B, C be a spherical triangle right-angled at C ; 
let a be assumed as the middle part ; then 

(90°— B) and b are the adjacent parts, 
and (90°— c), (90°— A), are the opposite parts. 
Then, by rule (1), 

R Xsin. a=:tan. (90°— B) tan. b ; 

=cot. B tan. b (a) 

By rule (2), 

R xsin. a = cos. (90°— A) cos. (90°— c); 

=sin. a sin. c (b) 

2d. Let b be the middle part ; then 

(90° —A) and a are the adjacent parts, 
and (90°— c) and (90°— B) are the opposite parts. 
Then, by rule (1), 

II sin. 5=tan. (90° — A) tan. a; 

=cot. A tan. a (c) 

Rule (2), 

R.sin. 5=cos.(90°— B) cos. (90°— c); 

=sin. B sin. c (d) 

3d. Let (90°— c) be the middle part; then 

(90°— A) and (90°— B) are the adjacent parts, 
and b and a are the opposite parts. 
Then, by rule (1), 

R.sin. (90°— c)=tan. (90°— A) tan. (90°— B) ; 

R cos. c = cot. A cot. B (e) 

Rule (2), R.sin. (90°— c)=cos. a cos. b ; 

R cos. c=cos. a cos. b . , (f) 



214 



SPHERICAL TRIGONOMETRY. 



4th. Let (90°— A) be the middle part; then 
(90°— c) and b are the adjacent parts, 
and (90° — B) and a are the opposite parts. 
Then, by rule (1), 

R sin. (90°— A)=tan. (90°— c) tan. b 
R cos. A=cot. c tan. b . . . 



(9) 



Rule (2), R sin. (90°-A)=cos. (90°-B) cos. a; 

R cos. A=sin. B cos. a . . (h) 
5th. Let (90°— B) be the middle part; then 
(90° — c) and a are the adjacent parts, 
and (90°— A) and b are the opposite parts. 
Then, by rule (1), 

R. sin. (90°— B)=tan.(90°— c) tan. a; 

R cos. B=tan. a cot. c . . . . (i) 
Rule (2), R.sin. (90°— B)=:COs. (90°— A) cos. b; 

.% R cos. B= sin. A cos. b . . (k) 
Collecting equations (a to k), we shall have 

R sin. a =cot. B tan. b (60) 

R sin. a =rsin. A sin. c (61) 

R sin. b =cot. A tan. a * (62) 

R sin. b =sin. B sin. c (63) 

R cos. c =cot. A cot. B (64) 

R cos. c =cos. a cos. b (65) 

R cos. A=tan. b cot. c (Q6) 

R cos. A = sin. B cos. a (67) 

R cos. B =:tan. a cot. c (68) 

R cos. B r^sin. A cos. b (69) 

It now remains to prove that these formulas are true. 

cos. c— cos. a cos. b 

Formula (3): cos. C= : -. — - 7 . 

v sin. a sin. b 

But when C=90°, cos. C=0 ; 

cos. c— cos. a cos. b 



therefore 



0=- 



sin. a sin. b 

or cos. c=cos. a cos. b, or R cos. c=cos. a cos. b* 
which is formula (65). 

Again, by (36), =-. — p, ; 

° J v J sin. c sin. C 

but when C=90°, sin. C=l ; therefore 

* We must multiply by E to make the equation homogeneous. 



SPHERICAL TRIGONOMETRY. 215 

sin. az=sin. A sin. c, 
which is formula (61). 

Also (37), ?^ = !^j? But if C=90°, sin. C=l ; 
v ' sm. c sin. C 

.\ sin. b — sin. B sin. c, 

which is formula (63). 

,:\ . cos. a— cos. b cos. c 

Also ( 1 ), COS. A = : — . 

v sm. o sm. c 

If we substitute for cos. c its value in (65), we shall have 
cos. a— cos. a cos. 2 b cos. a (1 — cos. 2 b) cos. a sin. 2 b 

COS.A= : — ; = . \ . '=- 7-; 

sin. b sm. c sm. 6 sm. c sm. e>. sin. c 

cos. a sin. 5 



sm. c 

Again, if we substitute for sin. c its value as found in 

(61), we have 

. cos. a sin. b cos. a sin. b sin. A 

cos. A= = : ; 

sin. a sm. a 

sin. A 
.-. sin. b—Qot. A tan. «, 
which is formula (62). 

. cos. a— cos. b cos. c 

Again, cos. A= : — -— . 

sm. b sm. c 

Substituting for cos. a its value in (65), we have 

cos. c 



cos. b 
cos. A= 



-cos. b cos. c 



sin. & sin. c 
cos. c sin. 5 



~ sin. c cos. 5 

=tan. b cot. c, 

which is formula (66). 

. # _. ,, cos. 5— cos. a cos. c 

Again (2), cos.B= : : . 

J sm. a sm. c 

Substituting for cos. c its value in (65), then 

_ cos. b— cos. b cos. 2 a 

COS. B = : : 

sm. a sm. c 
cos. b sin. a 
sin. c 



21G SPHERICAL TRIGONOMETRY. 

Now, substituting for sin. c its value in (63), we have 
' cos. b sin. a 

COS. B = ; ; ; 

sin. b 



sin. B 

.'•. sin. a— cot. B tan. b, 

which is formula (60). 

_ cos. b — cos. a cos. c 
Again, cos. B = : : . 

sin. a sin. c 

Substituting for cos. b its value in (65), we shall have 

cos. c 

cos. a cos. c 

_,, cos. a 

COS. B = ; ; 



sin. a sin. c 
cos. c sin. a 



*" sin. c cos. a 

,\ cos. B=tan. a cot. c, 

which is formula (68). 

. ■ fnK cos. A + cos. B cos. C 

Again, by (6), cos. «= : — — — - — - . 

6 ' J > " sm. B sin. C 

But when C=90°, cos. C=0, and sin. C = l ; 

cos. A 
.-. cos. a=— — — ; 
sm. B 

hence cos. A=:sin. B cos. a, 

which is formula (67). 

_ ", ' ' ft -i , cos. B+ cos. A cos. C 

And, by (7), cos. b— r 1 —. — : — ~ . 

' J v ; sm. A sin. C 

But when C= 90°, 

, cos. B 

cos. b=— t- ; 

sin. A 

.-. cos. B=sin. A cos. b, 

which is formula (69). 

• , , , A% cos. C+ cos. A cos. B 

Lastly, by (8), cos. c= : — r — : — ^ . 

■" J v n sm. A sm.B 

When C=90°, 

cos. A cos. B 

cos. C-- r — : — ^ ; 

sm. A sin. B 

.*. cos. e=cot. A cot. B, 

which is formula (64). 



SPHERICAL TRIGONOMETRY. 



217 



We have thus proved the results derived from Napier's 
Rules to be true, and may therefore, without hesitation, ap- 
ply those rules to the solution of the various problems in 
right-angled spherical triangles. 



SOLUTION OF EIGHT- ANGLED SPHERICAL 
TRIANGLES. 

In the solution of right-angled spherical triangles, we take 
the two given parts to find the other three parts in succes- 
sion. 

As the student sometimes finds a difficulty in ascertain- 
ing which is the middle part when he is applying Napier's 
Circular Parts, we shall give a few examples to illustrate 
the subject more fully. 

1°. When we have the hypothenuse, c, and the base, b, given, 
to find the other parts. 

To find the -angle A. 

Here the three parts lie together ; , b 

therefore the complement of the angle 
A is the middle part, and the extremes 
are adjacent. 

:. R.cos. A==tan. b cot. c. 

Now, as the middle part A is the 
unknown quantity, we begin the pro- 
portion with radius. 

As Radius : tan. b : : cot. c : cos. A ; 

whence the angle A is determined. 

To find the angle B. 

Here two of the parts, c and B, are 
adjacent to each other, but the side b is 
separated from both; hence b is the 
middle part, and the extremes are op- 
posite. 

.-. R.sin. b= sin. c sin. B ; 
Or, putting the equation into a pro- 
portion, 

K 






218 SPHERICAL TRIGONOMETRY. 

As sin. c : Radius : : sin. b : sin. B ; 

whence the angle B is determined. 

To find the side a. 

Here the two parts a and b are ad- 
jacent, because the right angle is alto- 
gether thrown out of consideration ; but 
the hypothenuse c is separated from 
both ; therefore the complement of c 
is the middle part, and the extremes 
are opposite. 

R.cos. c=cos. a. cos. b ; 
or, as cos. b : Radius : : cos. c : cos. a. 

Hence all the parts are determined from the two parts 
that were given. 

When a required part is determined from its sine, there 
will be two angles, both of which may be regarded as solu- 
tions, except when the ambiguity is removed by either of 
the following principles. 

(1.) In a right-angled spherical triangle, an angle and its 
opposite are always in the same quadrant / that is, they are ei- 
ther both less or both greater than 90°. 

(2.) When the two sides, including the right angle, are in the 
same quadrant, the hypothenuse is less than 90° ; and when the 
two sides are in different quadrants the hypothenuse is greater 
than 90°. 

EXAMPLES. 

1. In the right-angled triangle ABC are given the hy- 
pothenuse c=140°, and the perpendicular a=:20°, to find 
the other parts. Ans. A= 32° 8' 48". 

B=115°42' 23". 
5=144o36' 28". 

2. Given Ar=80° 10' 30", 5=155° 46' 42", to find the 
other parts. Ans. a= 67° 6' 52". 

5=110° 46' 20". 
B=155° 58' 24". 

3. Given a=116°, 5=16° to find the other parts. 

Ans.A= 97° 39' 24". 

B= 17° 41' 40". 

c=114° 55' 20". 



SPHERICAL TRIGONOMETRY. 219 

4. Given A=60° 47" 24", B=57° 16" 20", to find the 
other parts. Ans. a=54c° 32 / 33"". 

5=51° 43" 36"". 
c=68° 56" 29"". 



SOLUTION OF OBLIQUE-ANGLED SPHEEI- 
CAL TKIANGLES. 

The Problems and Theorems in the preceding pages on 
Spherical Trigonometry are sufficient to solve every trian- 
gle, or to obtain any part whatever of the triangle we may 
wish to obtain without going through the calculations of 
all the other parts. We shall only give the six different 
cases under which the spherical triangle may present itself. 

1°. When two sides and an angle opposite one of them are 
given. 

From Chap. II., th. 2, we derive the following 

RULE. 

As the sine of the side opposite the given angle : the sine of 
the side opposite the required angle : : the sine of the given an- 
gle : the sine of the required angle. 

We may then find the other angle from Prob. I., formula 
39, and the other side from Prob. II., formula 45. 

Or we may let fall a perpendicular from the unknown 
angle on the unknown side, and determine them by Napier's 
rules for circular parts. 

By the above rule we shall find two values for the an- 
gle supplementary to each other. If these two values, 
when used in formulas (39) and (45), cause neither of them 
to become negative, there will be two solutions. 

2°. When two angles and a side opposite one of them are 
given. 

Let A, B, and a be given. 

First find b by the equation 

sin. A sin. a 
sin. B — sin. V 
as before. Thus, 



220 SPHERICAL TRIGONOMETRY. 

As the sine of the angle opposite the given side : the sine of 
the angle opposite the required side : : the sine of the given side : 
the sine of the required side. 

We may then find C and c by using the same formulas 
as in the last case. 

We shall, as before, find two values for the side b sup- 
plementary to each other. 

If these two values, when substituted in formulas (39) and 
(45), cause neither to become negative, there will be two 
solutions. 

We may, as in the last case, let fall a perpendicular from 
the unknown angle on the unknown side, and find C and c 
by Napier's Circular Parts. 

That we may distinguish the trigonometric functions of 
parts greater than 90° from those less than 90°, we should 
carefully attend to the algebraic signs, using the plus and 
minus to the right of the logarithms. 

3°. When two sides and the included angle are given. 

Use the first of Napier's Analogies, p. 209. Then, to half 
the sum of the two angles add half their difference for the 
greater angle ; and from half the sum of the two angles sub- 
tract half their difference for the less angle. 

The third side can then be found by (1). 

4°. When two angles and the interjacent side are given. 

Use the second of Napier's Analogies^ Prob. II., p. 210. 
Then find each side separately by addition and subtraction, 
as in (3). 

5°. When the three sides are given. 

Any of the formulas from (14) to (22) will give the an- 
gles. . 



If we use tan. JA: 



/sin. (s—b) sin. (s—c) 
V sin. s. sin. (s— a) ' 



where s= %(a-\-b-{-c), we shall have 

log. tan. JA=| {log. sin. (s— 5)+ log. sin. (s—c) 
— log. sin. s— log. sin. (s— a). 
Now, since these formulas give the values of the half an- 
gles, there will be no ambiguity in this case. 
6°. When the three angles are given. 

Any of the formulas from (29) to (34) will give the 
sides. 

Using formula (26), we have 



SPHERICAL TRIGONOMETRY. 221 

2 log. cos. -|a=log. cos. (s / — B) + log. cos. (s'— C) 
—log. sin. B— log. sin. C, 
where s^^A+B+C). 

As in (5), there can be no ambiguity in the values of 
a, 5, and c. 

EXAMPLES. 

1. Given the angle B = 95° 30', the side c=64° 30', the 
side 5 = 100°, to find the other parts. 

Ans.A= 98° 32' 24". 
C = 65° 49' 25", 
a = 101° 55' 58". 

2. Given the angle A=61° 37' 52", B = 139° 54 7 34", 
and the side 5 = 150° 17' 28% to find the other parts. 

Ans. a=114° 26" 49". 
c= 82° S3' 31". 
C = 79° 10" 30". 

3. Given the side 5=99° 40" 48", the side c=100° 49' 
30", and the angle A = 65° 33' 10", to find the other 
parts. Ans. a=64° 23' 15". 

B=95° 38' 4". 
C = 97° 26' 29". 

4. Given the angle A=95° 38 / 4", the angle C = 97° 
26' 29", and the interjacent side 5 = 64° 23' 15", to find 
the other parts. Ans. a— 99° 40' 48" 

c=100°49'30 // . 
B= 65° 33' 10". 

5. Given the side a = 100°, the side 5=50°, and the side 
c = 60°, to find the angles. Ans. A=138° 15' 45". 

B= 31° 11" 14". 
0=, 35°49 / 58". 

6. Given the angle A = 115° 36' 45", the angle B = 
80° 19' 12", and the angle C=79° 10' 30", to find the 
sides. Ans. a=114° 26' 50". 

5= 84° 21" 56". 
c= 82° 33' 30" 



THE END. 






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